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

Mar 7, 2016

$= \int \frac{x + 1}{{x}^{2} + 6 x} d x$

#### Explanation:

$\int \frac{x + 1}{{x}^{2} + 6 x} d x$

Mar 7, 2016

$\frac{1}{6} \ln | x | + \frac{5}{6} \ln | x + 6 | + c$

#### Explanation:

First step is to factor the denominator.

${x}^{2} + 6 x = x \left(x + 6\right)$

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

thus: $\frac{x + 1}{x \left(x + 6\right)} = \frac{A}{x} + \frac{B}{x + 6}$

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 $\Rightarrow A = \frac{1}{6}$

let x = -6 in (1) : -5 = -6B $\Rightarrow B = \frac{5}{6}$

$\Rightarrow \frac{x + 1}{{x}^{2} + 6 x} = \frac{\frac{1}{6}}{x} + \frac{\frac{5}{6}}{x + 6}$

Integral can be written:

$\frac{1}{6} \int \frac{\mathrm{dx}}{x} + \frac{5}{6} \int \frac{\mathrm{dx}}{x + 6}$

$= \frac{5}{6} \ln | x | + \frac{5}{6} \ln | x + 6 | + c$