4.1 Integration By Parts
which can be rewritten
An Example
We know how to integrate ex really easily; on the other hand, we don’t
know, yet, how to integrate xex. But consider the same product:
. It
isn’t immediately obvious how relevant the derivative of this product is,
but let’s take its derivative anyway:
(
)=
+
∙
= ∙ −
+
,
∙
,
=
= ∙ −
or, in other words,
=
−
=
−
∙ ′
and now, we integrate this identity:
∙
=
∙
+
=
We have already seen how to integrate xex. The same procedure works
with sin x or cos x or any of the list of functions indicated above. Let’s
just try with sin x.
Start with the product uv, where both u and v are understood to be
functions of x. Since it is too much trouble (and too cluttered) to write
u(x) and v(x) all the time, we’re just going to write uv.
As before, we take the derivative of the product:
+
←
4.1.2 Examples: ex, sin x, cos x, sinh x, or cosh x multiplied by a
polynomial.
4.1.1 A formula for Integration By Parts
∙
=
Each of the arrows were taken to represent one of the terms in the
calculation. The first arrow, in blue, represents the original integral
you’re trying to find. The second arrow represents the u∙ v product
which is part of the answer. The last arrow, in green, is written
backward, to remind us that the formula requires a minus sign.
+ .
(I hope the color-coding helps to identify which terms has gone where.)
No one knows —at least I don’t— who invented this trick, but it is a very
commonly used trick, so much so that some people use it gratuitously
(though there’s obviously no harm in using this method even in cases
where a simpler method works just as well).
)=
→
↘
∙
=
which gives us
(
.
When the formula is used in practice, the calculation is usually set out in
a certain pattern which makes it easier to spot a mistake in the
calculations, as follows:
using the product rule.
Now, if we rewrite this equation in terms of integrals, we get
=
∙
∙
,
1
Example 1.
Integrate ∫ sin 4.1.3 Repeated application of By Parts
We’ve just seen the procedure for By Parts used twice in the same
problem. We took u to be the power of x, and each time we used it
resulted in (an integrated term, plus) a new integral which had one less
power of x. Since the other function was sine or cosine, we knew that if
we kept going, we would end up with no x’s at all, which would be totally
fine by us!
Repeated application of By Parts is also used in cases where the result
looks as if it is absolutely no improvement at all. In fact, in these
examples, we end up with the same thing we started out with! But with
a twist.
Suppose we want to integrate ∫
sin 3 . You can already
anticipate that, whichever factor we chose for u and whichever we chose
for v′, it very much looks as though we would be no better off. Luckily,
progress is made, though in a surprising fashion.
To begin with, we call the integral we’re trying to find by a name, e.g. I.
So,
.
Solution
In these problems, we must take u = x, or whatever polynomial is
present in the integral with the transcendental function (in this case,
sin x).
=
∙ sin = ∙ (−
) − (−
→
)∙
= sin
↘
= − cos + sin + .
=1
←
= − cos
[The arrows take up too much space on the page, so we’re going
to leave them out, and just put in a square grid instead.]
Example 2.
Integrate ∫
Solution
sin
= .
=−
2
Again, let’s take u = x , and v′=sinx.
=−
sin
cos − (−2 cos =−
3
)
=
= sin
′=2
= − cos
3
cos 3 −
−
cos 3 +
2
3
=−
=−
cos +
2 cos cos + 2 sin −
+
+
=2
2 sin =2
2
3
cos 3 =
= sin 3
′=2
1
= − cos 3
3
cos 3 What do we do now? We keep going. It is important to set u to be the
factor that was associated with the old u, and similarly for v:
=−
=−
sin 3
3
cos 3 +
= cos
=−
= sin
3
cos 3 +
2
9
2
3
sin 3 −
4
9
cos 3 =
2
=
sin 3
2
2
= cos 3
3
4
3
1
= sin 3
3
You’re probably not too surprised to see that we end up with a multiple
of our original integral!
But this is fine; we know algebra, so we can solve this as an algebraic
equation.
2
4
=−
cos 3 +
sin 3 −
3
9
9
To make life a little easier, let’s clear fractions. Multiply through by 9:
+ .
Important: The variable u is used twice, and the variable v is used twice
also. The second use of u is not the same as the first. (However, the
second use is related to the first use historically.) Similarly for v.
2
9 = −3
cos 3 + 2
sin 3 − 4 ,
so bringing the I over to the same side, we get
(2 sin 3 − 3 cos 3 ),
13 =
=
(
−
)+ .
Note how we throw in a +C at the end.
All the numbers in the result come out of the 2 and the 3 in the original
integral. 13, of course, is the square of 2 plus the square of 3.
A similar result is obtained when you integrate ∫
cos 3 , but
nobody memorizes these formulas; it is easy to obtain them when
needed.
This method also works for ∫ sec
.
4.1.4 Integrals of some transcendental functions
Sometimes it is convenient to take v′ to be just 1, which of course means
that v = x.
Integrate ∫ arctan .
= arctan
arctan
= arctan −
1+
= arctan − ln 1 +
=
1
1+
=1
=
+ .
Similarly for ln(x), arcsin (x) and arcsec (x).
3
5.
Exercises 2.3
Find the following integrals:
1.
(3 − )
7.
2.
3.
cos 4 sec
ln(
+ 1)
8.
4
cos(ln ) 4.2 Trigonometric Integrals
Take a few seconds to persuade yourself that this is correct. We are,
obviously, using the identity cos2 x + sin2 x = 1.
The parentheses can easily be expanded (or FOILed out, in high-school
jargon), and the integral becomes a pile of cosines, multiplied by a single sine:
Introduction
There are actually two families of methods we can use to find integrals
(antiderivatives) consisting of powers of trigonometric functions.
Obviously, we can easily find the following integrals:
1
1
sin , cos , sec 5 , csc
cot
,
2
2
and so on. We can also integrate sec x, sec3 x, and so on, using logarithms, etc.
We’re now going to capitalize on these possibilities, and simple substitution.
This is the first family of methods.
sin
∙ cos (− cos
1
18
1
=
sin
18
=
+
= cos + 3 cos
+ 3 cos
− cos
) cos
=
(−
+3
Mathematica gives
− 3 cos
+ cos
−3
)(− sin )
+
)
3
3
−
+
+ ,
7
9
11
13
which is horrible.
which a clever student can easily convert
back into a pile of cosines with appropriate powers.
=−
+
Even such an unlikely integral as ∫ cos √sin can be evaluated this
way; just replace the cubed cosine with (cosine) ∙ (cosine-squared), and
convert the squared cosine to 1 – sin2 x.
Mathematica gives
[ = sin
sin ∙ (1 − 3 cos
=
If we were to substitute = cos , the integral becomes
(remember, d = – sin x dx):
4.2.1 sines and cosines
Consider ∫ sin
∙ cos .
Think about this integral for a few minutes, and suggest a good choice for a
substitution. Remember, not only must you have an idea for the substitution,
you must also plan for the differential.
Well? If you suggested = sin x, you’d be in good company. (Obviously, you
can use any variable you wish; is just as good as any other one. Just avoid u
and v, because if you lose concentration for a second, you might think you’re
in the middle of a “By Parts” calculation.) So
sin
∙ cos
]
4.2.2 secants and tangents
Consider ∫ sec tan . For a really easy substitution, it would be
lovely if we had just sec
and then, any “polynomial” of tangents. But we
can get it to look like this! Replace sec with sec ∙ sec , and then
replace sec with (1 + tan ) ! Again, you must use an identity, and
simple rules of powers.
We end up with:
+
We can do actually more complicated integrals using a very simple trick.
Consider ∫ sin ∙ cos .
This next method works provided either the sine or the cosine has an odd
integer power. (For tangents and secants, different rules apply.)
Look at the sine factor. We rewrite it as
sin = sin ∙ sin = sin ∙ (1 − cos ) .
sec
∙ (1 + 2 tan
Now, use the substitution = tan , = sec
becomes ∫(1 + 2 + ) , which is easy.
7
)
+ tan
.
, and the integral
.
4.2.3 Products of Even Powers of Sine and Cosine
Consider ∫ sin ∙ cos . Since both powers are even, we can’t apply
the method from the beginning of this section; if we save either the sine or the
cosine, when we convert what’s left to the other function, we end up with
radicals! So we need to use a completely different approach.
The method is based on the identities
cos 2 = 2 cos
− 1,and cos 2 = 1 − 2 sin ,
from which we obtain
+
−
=
,and
=
.
Using these identities, our example is converted like this:
1 − cos 2 1 + cos 2
sin ∙ cos = ∙
2
2
1 − cos 2
=
.
4
As you can see, there is another even power of cosine present. The ¼ we can
easily integrate, but we must deal with the squared cosine:
1 + cos 4
1−
2
=
4
1 1
=
− cos 4 8 8
1
= − sin 4 + .
8 32
Note: Mathematica gives:
8
9.
Exercises 4.2
cos
Find the following:
1.
sin
cos
10.
3.
6.
cos
sin
cos
cos
1 + sin
11.
√sin sin
cos
28.
9
sec
26.
30.
sec
tan sec
∙ tan
42.
tan
(sec )
cot
45.
[Rewrite the integrand as
∙ csc ∙ cot , and
convert the extra cotangents into some powers of
csc t.]
36.
tan
sec
10
4.4.1 Radicals involving, (or fractional powers of,)
4.4 Trigonometric Substitutions
Let’s go back to the integral ∫
. While it is certainly nice that we can
√
force the integrand to stay defined by making the substitution
= 3 sin , there is a far greater advantage. If = 3 sin , then
√9 −
= 9 − (3 sin ) = √9 − 9 sin = √9 cos
= 3 cos , if is
chosen to lie in the interval − ≤ ≤ .
Introduction
This method is almost nothing more than plain old substitutions, except that
we do it just a little differently than simple algebraic substitutions.
Consider the integral
In general, if an integral contains a radical of the form √ − , the
suggested substitution is = sin , (or any other variable you like).
There are a number of other things that go with this substitution, which we
will show you in the examples, all of which help you replace the substitution
variable with x, or whatever you began with.
Example:
1
.
(4 − )
Following the general rule, we substitute
= 2 sin ,
= 2 cos .
.
√9 −
Normally we wouldn’t worry about things like the following, but can you
see that the integral doesn’t make sense outside the range –3 ≤ x ≤ 3?
For this reason, it makes sense to set x = 3 sin ( ), which would
guarantee that x stays within the range –3 ≤ x ≤ 3. Notice that this is a
sort of reverse substitution. Instead of setting some function of x to a
new variable, like w, here we’re setting just simply x equal to a
function of another variable.
Consider the integral
(
− 16)
−
.
At this point, we do something extra; we notice that sin
There isn’t an actual radical here, but you know that fractional powers are
powers of radicals. In this instance, we want the expression ( − 16) to
stay positive, otherwise the integral would not make sense. This requires
that
≥ 16. This can be ensured by taking x = 4 sec , for some variable ,
or even x = 4 csc . In either case, because of how secant and cosecant are
defined, we’re guaranteed that x will lie in the precise range of values that
makes the integrand defined.
Consider
1
,
√4 +
in this case x can be anything it likes! But it so happens that there is a choice
of x for which we can get rid of the radical!
= .
To make the process of replacing the ’s with x’s easier, we use
a right triangle into which we insert both x and , and 2.
2
Using our old high school Opposite-Adjacent-Hypotenuse
x
system, we decide that the Hypotenuse is going to be 2,
b
and because sine is opposite/hypotenuse, we figure
that the Opposite should be x. So we make the triangle
as
4-x 2
shown at right.
As we said, two of the sides (the hypotenuse and the opposite) were 2 and x.
Observe how we inserted the third side using Pythagoras’s theorem.
Now we’re in a position to write sine, cosine, tangent or anything at all using
various combinations of x’s.
We already showed you how to reduce the combination 4 − to 4 cos2 .
Replacing dx and the factor (4 − ) by their equivalents in terms of , we
get a new integral:
2 cos
1
=
sec .
8 cos
4
1
9
1
= arcsin
− 2
3
2
This is, of course, a snap to integrate. (Things usually don’t turn up this
brilliantly well.) The result is: tan + . Now, though we were prepared
9−
+ .
for anything, we only need to figure tan b in terms of x, which is
√
, so the final answer is
1
4 √4 −
4.4.2 Radicals involving
Consider the integral
+ .
.
√ −9
This does look similar to the previous example, but look at the radical closely:
the terms are reversed.
Again, we can not only substitute an expression for x which will guarantee
that the radical always is well-defined, but also get rid of the radical entirely.
The substitution is: = 3 sec ; = 3 sec ∙ tan ∙ , and again
there is a right triangle that helps us go back from the trig functions to the
algebraic expressions:
Example:
√9 −
The recommended substitution is
.
= 3 sin ,
= 3 cos sin
=
3
,
x
= 3 sec
a
3
−
9−x
2
= 3 sec
∙ tan
∙
9 sin
3 cos
= 9
sin
3cos 9
= (1 − cos 2 ) ,
2
using, of course, the little-known identity, the cosine double-angle/halfangle formula, which is a Calculus trade secret. Integrating:
9
1
=
− sin 2 + ,
2
2
√9 −
where, you must not forget, the variable represents an expression in x
that has to be replaced. To do this, recall that sin 2 = 2 sin cos .
From the little triangle, we see that
√9 −
cos =
=
.
ℎ
3
Continuing with the integral,
−
=
tan ,
sec
= ;
3
x
cos
3
In general, if the integral has the radical √
substitution is
= sec ; = sec ∙ tan ∙ ,
and, doing the algebra,
∙
x2 − 9
b
− 9 = 3 tan
So the integral becomes:
sec
x
,
−
=
.
b
(You need not memorize these; you have to use the
Trigonometry you know to derive these as you go along!)
a
Getting back to our example,
√
=
2
−9
9 sec
=
9 sec
∙ 3 sec ∙ tan
3 tan
9
= (sec tan
2
3
, the recommended
x
= ;cos
=
∙
+ ln|sec + tan |) + .
x 2 − a2
The integration part is essentially finished! But we still have to rewrite the
answer without any ’s. We have to use the triangle we set up along with the
substitution.
+ 25
25 √
2
1
=
2
=
9
√ −9
√ −9
=
∙
+ ln +
+
2 3
3
3
3
√ −9
9
= ∙
− +
+
− + − ln 3.
2
(The constant and (9/2) ln (3) can be replaced with another constant.)
tan ,
tan
2
x +a
x
=
sec
+
=
x
x
.
a
sec
=
Make sure that everything in the box above makes sense to you, from the
Trigonometry you know. Again, you ought not to memorize this; be prepared
to derive it.
Consider the integral ∫ √ + 25 .
Using the recommended substitution,
= 5 tan ,
= 5 sec
x√
x + 25
+ 25
tan
x
φ
+ 25 = 5 sec
x
= ,
5
= arctan
,
5
=
5 sec
+ 25
+
5
5
+
+ 25 + 25 ln
+ 25 +
+
−
=
sin
=
a
.
x
q
cos
∙ 5 sec
=
25 sec
−
= sec
sec ∙ tan
a2 − x 2
−
=
sec
=
x
∙
.
(This is most definitely a coincidence; the integral of secant-cubed is very
rare!)
3
+
=
sec
tan
= ,
x 2 − a2
,
b
tan
+
= tan ,
= sec = 5 tan , we get, using
2
,
+ tan |) +
+ 25
√
∙ + ln
5
5
−
= sin ,
= cos = ,
= arctan
q
+ ln|sec
tan
2
,
x
25
(sec
2
Summary:
4.4.3 Radicals of the form
+
These sorts of integrals can be rewritten using a substitution so that the
radical is completely removed, leaving only a formula without any radicals.
The substitution is
=
=
a
x 2 + a2
,
q
x
a
4
3.
Exercises 4.4 [Find, and do, extra problems wherever you can!]
− 1
1.
(4 −
)
5.
√9 −
[1 bonus point if you do this two different ways.]
2.
(9 +
)
5
7.
√9
−1
( + 1) √
+4
√
1
8.
1
9
+2 −5
10.
1
√9 −
[Hint: Let y = x+1]
6
4.3 Partial Fractions
where A, B, C are constants, and we must find constants that will make
the two sides of the equation equal for all values of x. (The two sides are
probably going to be equal for almost any choice of A, B, C for a
particular value of x, but that’s useless to us.) This is very important.
This equation actually has to be solved identically in x.
First, we clear fractions. This is a standard procedure for getting rid of
fractions in an equation: just multiply through by the Lowest Common
Denominator of both sides. In all these cases, this lowest common
denominator (or LCM) will be the denominator of the original fraction;
in the present problem ( − 2)( + 1)( − 3). Doing this, we get:
− 4 = ( + 1)( − 3) + ( − 2)( − 3) + ( − 2)( + 1).
To find A, B, C, there are at least two methods. We use the most difficult
method, because it always works. (There is an easier method that works
some of the time.)
Now, the easiest thing to do, especially if you have trouble with algebra,
is to multiply out the parentheses:
− 4 = ( − 2 − 3) + ( − 5 + 6) + ( − − 2).
Combining the groups on the right, you know we’re going to get a
polynomial, and this polynomial has to be identical to the polynomial on
the left. This means,
The “x2” terms on both sides must match exactly:
0 = A + B + C,
The “x” terms on both sides must match exactly:
1 = 2A – 5B  C,
The constant terms on both sides must match exactly:
4 = 3A + 6B  2C.
Here’s a convenient way of solving these sorts of equations.
We distil the information in these 3 equations into what is called a
matrix:
Introduction
The method of partial fractions is really just an algebraic method for
splitting up fractions, which of course makes fractions easier to
integrate. [Don't confuse this method with By Parts!!!]
We’ll begin with easy fractional decompositions, but remember: as the
technique gets more sophisticated, you cannot simply fall back on this
first procedure. This method applies only to very easy problems.
In every case, the fraction must have a numerator of smaller degree
than the denominator. There are very good reasons for this, which will
make sense only after you see a few examples.
4.3.1 A very easy special case: A fraction with distinct linear
factors
Consider the integral
−4
.
( − 2)( + 1)( − 3)
The integrand looks a lot more innocent than lots of things we’ve been
integrating already, but you must agree that nothing we’ve described so
far is going to work. (The only thing that might have worked would have
been some sort of natural logarithm, but a quick check would dash your
hopes there.)
But look at the integrand. Having checked first that the numerator is
definitely of smaller degree than the denominator, a little reflection will
persuade you that if
−4
( − 2)( + 1)( − 3)
was obtained by combining several simple fractions, those fractions must
have had denominators of (x – 2), (x + 1), and (x – 3). So, the simplest
procedure in this group of methods is to find the fractions that will work.
We set
−4
=
+
+
,
( − 2)( + 1)( − 3)
−2
+1
−3
A
B
C
 1 1 1
 2 5 1

 3 6 2
9
0
1
4
(The A, B, C on top is merely decoration.) The numbers we had on the
left sides of the three equations really represent the formula x – 4. No x
squared, one x, and a 4. The numbers under the A represent the
polynomial ( − 2 − 3), the numbers under B represent ( − 5 +
6), and so on.
The usual method of solving equations of this sort is elimination through
adding and subtracting. We do the same, but the matrix method involves
less writing, actually.
The method proceeds by adding one row (which is one equation, really,)
to another row, or sometimes a multiple of one row to another, to create
as many zeros as possible. The row that is added is called the pivot row.
The row that it is added to is called the target row.
Let’s try it out. The top row has all 1’s, so it is the most convenient to
use; let’s add two times the top row to row 2:
 3 3 3 0
0 3 1 1


3 0 0 2
Now, we subtract the last row from the top row:
0 3 3 2
0 3 1 1


3 0 0 2
Well, what do you think might help next? Notice that if we add the
middle row to the top row, we introduce another zero in the top row:
0 0 4 1
 0 3 1 1 


3 0 0 2
 1 1 1 0
 0 3 1 1


 3 6 2 4
Finally, we can add a quarter of the top row to the middle row, to get:
0 0 4 1

5 
0 3 0
4


2
3 0 0
Notice that, though we multiplied row 1 by two, we keep it unchanged.
Next, we can add two times row 2 to the third row, getting rid of the 6
and the -2:
 1 1 1 0
 0  3 1 1


 3 0 0 2
What has been described here is not the most systematic procedure, but
the one that requires the least amount of technique. Those of you who
are learning a more careful version of this method in other classes must
continue to follow the procedure taught there!
Now, since we have only one non-zero entry in each row, and each nonzero entry in a different column, we rewrite the equations in terms of A,
B and C to get: 3A = 2; 3B = ¾, and 4C = 1. So, A = 2/3, B = ¼, and
C = ¼ also.
This means that our integral can be rewritten:
−5
−1
2
3 + 12 + 4
−2
+1
−3
Another sort of operation we’re allowed is to multiply any row by any
number (except zero!). In this case, we can multiply the last row through
by -1, which will be convenient.
1 1 1 0
0 3 1 1


3 0 0 2
Next, we multiply row 1 by three:
1
= 2 3 ln| − 2| − 5 12 ln| + 1| − ln| − 3| + .
4
10
+8
.
( − 2)
Because of the repeated factor, here’s the partial fraction structure that
we “try”:
+8
= +
+
,
( − 2)
− 2 ( − 2)
where, of course, no surprises, we have to find A, B, and C.
Clearing fractions,
+ 8 = ( − 2) + ( − 2) + .
As before, you can certainly multiply out the right side. But it is a lot more
convenient to keep it the way it is, and simply think what the coefficients
of the right side will be.
Just in case some students were wondering whether there isn’t another
way to do this problem, there is.
We have:
=0
+ +
2 + 5 + = −1
3 − 6 + 2 = 4,
(where we have neatened the equations just a little bit).
If you happened to notice (and who can notice these things?) that when
the first two equations are subtracted from the last, we get:
−12 = 5,
from where we get B right away: B = –5/12.
The usual thing anyone does, at this point, is to rewrite equations 1 and
2, for instance, with the known value for B:
5
+ = ⋯ (1),
12
2 + = −1 + , or
The highest power of x present in the identity is x2.
Coefficient of x2 on the Left Side =0.
Coefficient of x2 on the Right Side = A + B. (The C term has no x2s.)
Next we do x. Coefficient of x on the Left Side = 1.
Coefficient of x on the Right Side = – 4A – 2B + C. This is the only tricky
one to figure out; if you can see that this is true, you could probably do
this sort of thing anytime. Basically, you’re foiling the terms out in your
head, and picking out just the x terms in each group.
Finally, the constants. On the Left Side: 8;
On the Right Side: 4A. Do you see this? No constant terms appear with
either B or C. So the equations are:
+ = 0,
−4 − 2 + = 1,
4 = 8.
Of course A must be 2, from the last equation; backing up to the first
equation, B = –2; plugging these in for A and B, C = 5.
13
⋯ (2).
12
Now getting A and C is easy; for instance just subtract (1) from (2), and
we get A = 2/3, so that B has to be ¼.
As you can see, the matrix method leads to the answer more steadily,
especially if you’re not afraid to use fractions all through the problem.
2 + =
4.3.2 Repeated linear roots
Consider
So, our integrand can be split up into
+8
.
−4 +4
The denominator does have higher degree than the top, so we can
proceed. In fact, notice that the denominator can be factored. (Don’t
waste time factoring the top, unless there’s going to be some cancelling.)
So the integral becomes:
Performing the integration,
+8
=
( − 2)
2
= 2 ln − 2 ln| − 2| −
11
−
−
+(
)
.
2
5
+
− 2 ( − 2)
5
+ .
−2
(Make sure you understand how the last term is found.)
If A and k have the same sign, then we have what is called an irreducible
quadratic polynomial.
For the simple reason that we don’t want students racking their brains
over whether or not some factor is irreducible, we give very obviously
irreducible factors in Calc 2. For instance, both these polynomials are
irreducible:
4 + 4 + 5,4 + 9
but a student is far less likely to spot the one on the left as irreducible,
than the one on the right.
Consider the integral
−8 −4
.
( − 4 )( + 4)
The numerator has smaller degree than the denominator, so we can
proceed to split into partial fractions. The big new piece of information
is that each irreducible factor must have a numerator of the form Ax + B.
(Of course you have to use different unknowns for each factor.):
−8 −4
+
= +
+
.
( − 4 )( + 4)
−4
+4
Clearing fractions:
− 8 − 4 = ( − 4)( + 4) + ( + 4) + ( + )( − 4 ).
Now we compare the coefficients of various powers:
x3: 0 = A + B + C;
x2: 1 = – 4A – 4C + D; x: –8 = 4A + 4B – 4D,
Constants: –4 = –16A.
A = 1/4. We can rewrite the remaining equations as
There are two more sorts of factors in a denominator that must be dealt
with, and then you must also learn what to do with a fraction where the
numerator has higher degree than the denominator.
Let’s do the last technique right away.
4.3.4 Improper Fractions; that is, numerator higher degree than
denominator
Consider
4 − 16 + 16 + + 8
.
−4 +4
We just perform long division. You’re expected to know how to do this:
4 − 16 + 16 + + 8
+8
= 4 +
.
( − 2)
−4 +4
So
4 − 16 + 16 + + 8
+8
=
4 +
( − 2)
−4 +4
5
= 2 + 2 ln − 2 ln| − 2| −
+ .
−2
4.3.5 Irreducible Quadratic Factors
You must remember from earlier courses that a quadratic of the form
+
+
can always be written in a form with the completed square as follows:
( − ℎ) + .
If A and k have opposite signs, then this expression can be factored into
linear factors, even if there might be some ugly radicals involved. So if
the denominator contains something like 3 + 4 − 5, in principle it
can be factored; in this case the factors will look like 3 + 2 + √19 +
2 − √19 , which of course will be tedious to break into partial fractions,
but still possible.
B + C = –1/4,
4C – D = –2,
B – D = –9/4.
Combining the two equations that
have B in them, we get C + D = 2.
Now adding to the remaining
equation, we get 5C = 0, so C =0.
So, what does the integral look like now?
−8 −4
( − 4 )( + 4)
1
1
2
=
−
+
(
)
4
4 −4
+4
1
1
2
=
− ln| − 4| + arctan
+ .
4 ln
4
2
2
12
This gives B = –1/4;
D = 2.
Exercises:
1
3
6
1
+2 −3
+1
( − 1)
1
( − 1)
10
( − 1)(
13
+ 1)
13
18
(
−1
+ 1)
[Ask me for a Hint. It has to do with multiplying and dividing by ex.]
14
(
+ 1)
31
14
+1
+