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 #