5 Steps to a 5 AP Calculus BC 2019

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

1. Evaluate

∫

1

x^2

dx.

Answer:Rewrite as

∫

x−^2 dx=

x−^1

− 1

+C=−

1

x

+C.

2. Evaluate

∫

x^3 − 1

x

dx.

Answer:Rewrite as

∫ (

x^2 −

1

x

)

dx=

x^3

3

−ln|x|+C.

3. 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.

4. Evaluate

∫

sinxdx.

Answer:−cosx+C.