Integration 223
Example 2
∫
x^5 + 2 x^2 + 1
x^3 −x
dx
Step 1: Use long division to rewrite
∫
x^5 + 2 x^2 + 1
x^3 −x
dx=
∫ (
x^2 + 1 +
2 x^2 +x+ 1
x^3 −x
)
dx=
∫
(x^2 +1)dx+
∫
2 x^2 +x+ 1
x^3 −x
dx.
Step 2: Factor the denominator:
2 x^2 +x+ 1
x^3 −x
=
2 x^2 +x+ 1
x(x+1)(x−1)
Step 3: Let A, B, and C represent the numerators of the partial fractions.
2 x^2 +x+ 1
x(x+1)(x−1)
=
A
x
+
B
x+ 1
+
C
x− 1
Step 4: 2 x^2 +x+ 1 =A(x+1)(x−1)+Bx(x−1)+Cx(x+1), therefore,Ax^2 +Bx^2 +Cx^2 =
2 x^2 ,Cx−Bx=x, and−A=1. Solving givesA=−1,B=1, andC=2.
Step 5:
∫
x^5 + 2 x^2 + 1
x^3 −x
dx=
∫
(x^2 +1)dx+
∫
− 1
x
dx+
∫
1
x+ 1
dx+
∫
2
x− 1
dx
=
x^3
3
+x−lnx+ln (x+1)+2ln(x−1)+C
10.4 Rapid Review
- Evaluate
∫
1
x^2
dx.
Answer:Rewrite as
∫
x−^2 dx=
x−^1
− 1
+C=−
1
x
+C.
- Evaluate
∫
x^3 − 1
x
dx.
Answer:Rewrite as
∫ (
x^2 −
1
x
)
dx=
x^3
3
−ln|x|+C.
- Evaluate
∫
x
√
x^2 − 1 dx.
Answer:Rewrite as
∫
x(x^2 −1)^1 /^2 dx. Letu=x^2 −1.
Thus,
du
2
=xdx⇒
1
2
∫
u^1 /^2 du=
1 u^3 /^2
23 /^2
+C=
1
3
(x^2 −1)^3 /^2 +C.
- Evaluate
∫
sinxdx.
Answer:−cosx+C.