### SOLUTIONS TO INTEGRATION BY PARTIAL FRACTIONS

SOLUTION 9 : Integrate . Decompose into partial fractions (There is a repeated linear factor !), getting

(After getting a common denominator, adding fractions, and equating numerators, it follows that ;
let ;
let ;
let ;
it follows that and .)

.

SOLUTION 10 : Integrate . Factor and decompose into partial fractions (There is a repeated linear factor !), getting

(After getting a common denominator, adding fractions, and equating numerators, it follows that

;

let ;
let ;
let ;
let
.)

.

SOLUTION 11 : Integrate . Factor and decompose into partial fractions (There are two repeated linear factors !), getting

(After getting a common denominator, adding fractions, and equating numerators, it follows that

;

let ;
let ;
let
;
let
;
it follows that and .)

.

SOLUTION 12 : Integrate . Use the method of u-substitution first. Let

so that

.

Substitute into the original problem, replacing all forms of , getting

(Factor and decompose the function into partial fractions. There is a repeated linear factor !)

(After getting a common denominator, adding fractions, and equating numerators, it follows that ;
let ;
let ;
let .)

(Recall that .)

.

SOLUTION 13 : Integrate . Because the degree of the numerator is not less than the degree of the denominator, we must first do polynomial division. Then factor and decompose into partial fractions, getting

(After getting a common denominator, adding fractions, and equating numerators, it follows that

;

let ;
let ;
let .)

.

SOLUTION 14 : Integrate . Use the method of u-substitution first. Let

so that

.

Substitute into the original problem, replacing all forms of , getting

(Decompose into partial fractions.)

(After getting a common denominator, adding fractions, and equating numerators, it follows that ;
let ;
let .)

(Recall that .)

.