Integral Calculus
Grinshpan
1.
Two examples on partial fractions
x2
.
Find the indefinite integral of 2
x − 3x + 2
SOLUTION
x2
(x2 − 3x + 2) + (3x − 2)
3x − 2
=
=1+ 2
2 − 3x + 2
2 − 3x + 2
x
x
x
3x + 2}
|
{z
}
| − {z
improper...
proper!
Decompose the proper part:
3x − 2
3x − 2
P
Q
=
=
+
− 3x + 2
(x − 1)(x − 2)
x−1 x−2
3x − 2 = P (x − 2) + Q(x − 1)
Clear denominators:
1 = −P
Set x = 1:
Set x = 2:
4=Q
Z
Integrate:
2.
x2
x2
dx =
x2 − 3x + 2
Z 1
4
1−
+
dx = x − ln |x − 1| + 4 ln |x − 2| + C
x−1 x−2
Find the partial fraction decomposition for
1 − x2
. Find the integral.
x3 (x2 + 2)
SOLUTION
Write the form of the decomposition:
1 − x2
A
B
C
Dx + E
= + 2+ 3+ 2
.
3
2
x (x + 2)
x
x
x
x +2
Multiply both sides by x3 (x2 + 2): 1 − x2 = Ax2 (x2 + 2) + Bx(x2 + 2) + C(x2 + 2) + (Dx + E)x3 .
Arrange terms by the powers of x: −x2 + 1 = (A + D)x4 + (B + E)x3 + (2A + C)x2 + 2Bx + 2C.
Compare the coefficients
x0 :
1 = 2C
1
0 = 2B
x :
Conclude that C =
1
,
2
B=0,
Obtain the decomposition:
x2 :
−1 = 2A + C
3
x :
0=B+E
x4 :
0=A+D
A=−
3
,
4
E=0,
D=
3
.
4
1 − x2
3/4 1/2
3x/4
=−
+ 3 + 2
.
3
2
x (x + 2)
x
x
x +2
At this point, integration does not take much effort:
Z
1 − x2
3
1
3
dx = − ln |x| − x−2 + ln(x2 + 2) + arbitrary constant.
3
2
x (x + 2)
4
4
8