204 CHAPTER 8. METHODS FOR FINDING ANTIDERIVATIVES

(a)∫ √

4x2 +16x+15dx

(b)∫ √

x2 +6xdx

(c)∫ 3√

−32−9x2−36xdx

(d)∫ 3√

−5−x2−6xdx

(e)∫ 1√

9−16x2−32xdx

(f)∫ √

4x2 +16x+7dx

5. Find∫

x5√

1+ x4dx.

6. Find∫ x√

1−x4dx. Hint: Try x2 = sin(u).

8.7 Partial FractionsThe main technique for finding antiderivatives in the case f (x) = p(x)

q(x) for p and q poly-nomials is the technique of partial fractions. Before presenting this technique, a few moreexamples are presented. These examples are typical of the kind of thing you end up doingafter you have found the partial fractions expansion.

Example 8.7.1 Find∫ 1

x2+2x+2 dx.

To do this, complete the square in the denominator to write∫ 1x2 +2x+2

dx =∫ 1

(x+1)2 +1dx

Now change the variable, letting u = x+1, so that du = dx. Then the last indefinite integralreduces to ∫ 1

x2 +2x+2dx = arctan(x+1)+C.

Example 8.7.2 Find∫ 1

3x+5 dx.

Let u = 3x+5 so du = 3dx. Then you obtain∫ 1

3x+5 dx = 13 ln |3x+5|+C.

Example 8.7.3 Find∫ 3x+2

x2+x+1 dx.

First complete the square in the denominator.∫ 3x+2x2 + x+1

dx =∫ 3x+2

x2 + x+ 14 +

34

dx =∫ 3x+2(

x+ 12

)2+ 3

4

dx.

Now let(x+ 1

2

)2= 3

4 u2 so that x+ 12 =

√3

2 u. Therefore, dx =√

32 du and changing the

variable, one obtains

=

√3

2

(2√

3∫ u

u2 +1du− 2

3

∫ 1u2 +1

du)=

32

ln(u2 +1

)−

√3

3arctanu+C

Therefore,∫ 3x+2

x2+x+1 dx =

32

ln

((2√3

(x+

12

))2

+1

)−

√3

3arctan

(2√3

(x+

12

))+C.