1.
Z
2−x
dx
x2 + 5x
Solution: We did this one in section. The answer is
2
7
ln |x| − ln |x + 5| + C
5
5
2.
Z
x4 + x3 + x2 + 1
dx
x2 + x − 2
Solution: Since the degree of the numerator is greater than the degree of the denominator, first we need to do polynomial long division:
x2 + x − 2
so that
x2
+3
4
3
2
x +x +x
+1
− x4 − x3 + 2x2
3x2
+1
− 3x2 − 3x + 6
− 3x + 7
−3x + 7
x4 + x3 + x2 + 1
= x2 + 3 + 2
2
x +x−2
x +x−2
Since x2 + x − 2 = (x + 2)(x − 1), the fraction may be decomposed. We worked through
this in section and got
−3x + 7
4
13
=
−
x2 + x − 2
3(x − 1) 3(x + 2)
So
Z
x4 + x 3 + x2 + 1
dx =
x2 + x − 2
Z
Z
Z
4
13
x + 3dx +
dx −
dx
3(x − 1)
3(x + 2)
x3
4
13
=
+ 3x + ln |x − 1| −
ln |x + 2| + C
3
3
3
2
1
3.
Z
x3 − x2 + 3x − 3
dx
x4 − 3x3 − x2 + 3x
Solution: The denominator factors as
x4 − 3x3 − x2 + 3x = x(x − 1)(x + 1)(x − 3)
(If you’re not comfortable with factoring polynomials, take a look at http://www.purplemath.
com/modules/solvpoly.htm)
These are distinct linear factors, so the partial fraction decomposition is straightforward.
Solve for the coefficients to get
x3 − x2 + 3x − 3
1
1
1
=− +
+
4
3
2
x − 3x − x + 3x
x x+1 x−3
(Notice that the coefficient over (x − 1) is zero. This is because it turns out that the
numerator is also divisible by (x − 1), so the (x − 1) factors cancel).
Thus,
Z
x3 − x2 + 3x − 3
dx = − ln |x| + ln |x + 1| + ln |x − 3| + C
x4 − 3x3 − x2 + 3x
2
? 4.
Z
x4
−
4x3
dx
+ 6x2 − 8x + 8
Solution: The denominator factors as
x4 − 4x3 + 6x2 − 8x + 8 = (x − 2)2 (x2 + 2)
Note that x2 + 2 is irreducible since it does not have any roots. Now, the partial fraction
decomposition should look like
A
B
1
Cx + D
=
+
+
x4 − 4x3 + 6x2 − 8x + 8
x − 2 (x − 2)2
x2 + 2
Multiply through by x4 − 4x3 + 6x2 − 8x + 8 to get
1 = A(x − 2)(x2 + 2) + B(x2 + 2) + (Cx + D)(x − 2)2
Multiplying out the right hand side and grouping together like terms, we find
1 = (A + C)x3 + (−2A + B − 4C + D)x2 + (2A + 4C − 4D)x + (−4A + 2B + 4D)
This gives the system of equations
A+C
−2A + B − 4C + D
2A + 4C − 4D
−4A + 2B + 4D
=0
=0
=0
=1
Solve this system by any method you like (substitution will work) to get
1
A=− ,
9
1
B= ,
6
1
C= ,
9
D=
1
18
Thus,
Z
dx
1
=−
4
3
2
x − 4x + 6x − 8x + 8
9
Z
dx
1
+
x−2 6
Z
dx
1
+
2
(x − 2)
9
Z
2x + 1
dx
x2 + 2
The first two integrals are straightforward: (we omit the constant until the end)
Z
1
dx
1
−
= − ln |x − 2|
9
x−2
9
Z
1
dx
1
=−
2
6
(x − 2)
6(x − 2)
The third integral is trickier. First, distribute the integral over the sum to get
Z
Z
1
2x
1
1
dx
+
dx
9
x2 + 2
9
x2 + 1
3
Now, the first integral can be evaluated by substituting u = x2 + 2 with du = 2xdx to
get
Z
Z
2x
1
du
1
1
1
dx =
= ln |u| = ln |x2 + 2|
2
9
x +2
9
u
9
9
To evaluate the second integral, rewrite it as
Z
Z
1
1
1
1
dx
=
dx
2
9
x2 + 2
18
x
√
+1
x
√
√
Now, subsitute u = x/ 2 with du = 1/ 2 to get
√ Z
√
√
√
du
2
2
2
=
arctan(u)
=
arctan(x/
2)
18
u2 + 1
18
18
Putting all of the pieces together, we get
Z
√
√
dx
1
1
2
1
2
ln
|x−2|−
+
ln
|x
+2|+
arctan(x/
=
−
2)+C
x4 − 4x3 + 6x2 − 8x + 8
9
6(x − 2) 9
18
4
? 5.
Z
π/6
π/3
cos(x)
dx
sin2 (x) + 41
Solution: Substitute u = sin(x)
√ so that du = cos(x)dx. To get at the limits of integration, notice that sin(π/3) = 3/2 and sin(π/6) = 12 . Thus, the integral becomes
Z
1/2
√
3/2
du
u2 +
1
4
Now, multiply numerator and denominator by 4 (similar to the previous problem) to get
Z
1/2
√
3/2
4du
=4
4u2 + 1
Z
1/2
√
3/2
du
(2u)2 + 1
Finally, subsitute v = 2u with 2v = 2du and limits of integration
Z
2 √1
= 2 arctan(v)|√31
3
dv
v 2 +1
√
= 2 arctan(1) − 2 arctan( 3)
π
π
=2· −2·
4
3
π 2π
= −
2
3
π
=−
6
5
√
3 and 1 to get
6.
Z
dx
1
1 + x3
Solution: This is a substitution problem, but instead of trying to find du inside of the
integral in terms of x, we will solve for x in terms of u, find dx, and plug in. We want to
get rid of that cube root, so substitute
1
u = x3
Then
u3 = x
so
dx = 3u2 du
Plugging in, we get
Z
Z
dx
1
=
3u2
du
1+u
1 + x3
Since the degree of the numerator is greater than the degree of the denominator, we can
do polynomial long division to simplify this.
3x − 3
x+1
3x2
− 3x2 − 3x
− 3x
3x + 3
3
so that
Z
3u2
=
1+u
Z
Z
3u − 3du +
3
1+u
3
= u2 − 3u + 3 ln |1 + u|
2
1
1
3 2
= x 3 − 3x 3 + 3 ln |1 + x 3 |
2
7.
Z
3x + 2
√
dx
x−9
Solution:
http://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/powersubsoldirectory/PowerSubSol.
html#SOLUTION%205
6
? 8.
Z
√
x+1
√ √
dx
x( 3 x + 2)
Solution:
http://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/powersubsoldirectory/PowerSubSol2.
html#SOLUTION%207
7
9.
Z
x sin x cos xdx
Solution:
http://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/intbypartssoldirectory/
IntByPartsSol2.html#SOLUTION%2014
? 10.
Z √
1 + e2x dx
Solution: This one is pretty tricky. We substitute
√
u = 1 + e2x
Then
e2x = u2 − 1
and taking derivatives of both sides we get
2e2x dx = 2u du
so that
2(u2 − 1)dx = 2u du
and
dx =
u2
u
du
−1
Thus,
Z √
Z
1+
e2x dx
u·
=
u2
u
du
−1
u2
u2 − 1
Z 2
u −1+1
=
u2 − 1
Z
1
= 1+ 2
u −1
Z
=
For the second term, we use the partial fraction decomposition
u2
1
1
1 1
1 1
=
=−
+
−1
(u − 1)(u + 1)
2u−1 2u+1
so that our integral becomes
Z
Z
Z
1
du
1
du
1
1
du −
+
= u − ln |u − 1| + ln |u + 1| + C
2
u−1 2
u+1
2
2
√
√
√
1
1
= 1 + e2x − ln | 1 + e2x − 1| + ln | 1 + e2x + 1| + C
2
2
8
11.
Z
dx
+1
ex
Solution: Substitute u = ex + 1. Then du = ex dx = (u − 1)dx so that
du
u−1
dx =
Thus
Z
dx
=
x
e +1
Z
du
u(u − 1)
The partial fraction decomposition is
1
1
1
=− +
u(u − 1)
u u−1
and
Z
Z
Z
du
1
1
=−
du +
du
u(u − 1)
u
u−1
= − ln |u| + ln |u − 1| + C
= − ln |ex + 1| + ln |ex | + C
9
12.
Solution: Recall that
Z
d
(sec(x))
dx
sec5 x tan3 xdx
= sec(x) tan(x). We can write
sec5 x tan3 x = sec4 x tan2 x(sec x tan x)
The substitution u = sec x would work if we could write tan2 x as some expression in
terms of sec x. Fortunately, we do know how to do this! We have the trig identity
tan2 x = sec2 x − 1. Thus
Z
Z
5
3
sec x tan xdx = sec4 x tan2 x(sec x tan x)dx
Z
= sec4 x(sec2 x − 1)(sec x tan x)dx
Z
= u4 (u2 − 1)du
Z
= u6 − u4 du
u7 u5
−
+C
7
5
sec7 x sec5 x
=
−
+C
7
5
=
10