Question

How do you integrate `int (x+1)/(x^2 + 6x)` using partial fractions?

Answer 1

`=int (x+1)/(x^2+6x)d x`

Explanation:

`int (x+1)/(x^2+6x)d x`

Answer 2

`1/6ln|x| + 5/6ln|x+6| + c`

Explanation:

First step is to factor the denominator.

`x^2 + 6x = x(x+6)`

Since these factors are linear , the numerators of the partial fractions will be constants , say A and B.

thus: `(x+1)/(x(x+6)) = A/x + B/(x+6)`

multiply through by x(x+6)

x+ 1 = A(x+6) + Bx ......................................(1)

The aim now is to find the value of A and B. Note that if x = 0. the term with B will be zero and if x = -6 the term with A will be zero.

let x = 0 in (1) : 1 = 6A `rArr A = 1/6`

let x = -6 in (1) : -5 = -6B `rArr B = 5/6`

`rArr (x+1)/(x^2+6x) = (1/6)/x + (5/6)/(x+6)`

Integral can be written:

`1/6int(dx)/x + 5/6int(dx)/(x+6)`

`= 5/6ln|x| + 5/6ln|x+6| + c`