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

2 Answers
Mar 7, 2016

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

Explanation:

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

Mar 7, 2016

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