How do you integrate int (x+1)/(x^2 + 6x) using partial fractions?
2 Answers
Mar 7, 2016
Explanation:
Mar 7, 2016
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