# How do you find int 10/((x-1)(x^2+9)) dx using partial fractions?

Oct 29, 2015

$\ln | x - 1 | - \frac{1}{2} \ln | {x}^{2} + 9 | + \frac{1}{3} {\tan}^{- 1} \left(\frac{x}{3}\right) + C$

#### Explanation:

Since the denominator consists of 1 linear factor and 1 irreducible quadratic factor, we may use the method of partial fractions to rewrite it as :

$\frac{10}{\left(x - 1\right) \left({x}^{2} + 9\right)} = \frac{A}{x - 1} + \frac{B x + C}{{x}^{2} + 9}$

$= \frac{A \left({x}^{2} + 9\right) + \left(B x + C\right) \left(x - 1\right)}{\left(x - 1\right) \left({x}^{2} + 9\right)}$

$\therefore 10 = A {x}^{2} + 9 A + B {x}^{2} - B x + C x - C$

$= \left(A + B\right) {x}^{2} - \left(B - C\right) x + \left(9 A - C\right)$

Now comparing terms we get that

$A + B = 0$
$C - B = 0$
$9 A - C = 10$

Solving this linear system of equations yields :

$A = 1 , B = - 1 , C = - 1$

Therefore the original integral may be written and solved as

$\int \frac{10}{\left(x - 1\right) \left({x}^{2} + 9\right)} \mathrm{dx} = \int \frac{1}{x - 1} \mathrm{dx} - \int \frac{x}{{x}^{2} + 9} \mathrm{dx} - \int \frac{1}{{x}^{2} + 9} \mathrm{dx}$

$= \ln | x - 1 | - \frac{1}{2} \ln | {x}^{2} + 9 | + \frac{1}{3} {\tan}^{- 1} \left(\frac{x}{3}\right) + C$