# How do you integrate 10/[(x-1)(x^2+9)] dx?

Jun 17, 2015

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

#### Explanation:

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

So we want: $A {x}^{2} + 9 A + B {x}^{2} - B x + C x - C = 10$
Hence:
$A + B = 0$
$- B + C = 0$
$9 A - C = 10$

The second equation gives us: $B = C$, so
$A + C = 0$
$9 A - C = 10$, adding gets us:

$10 A = 10$, so $A = 1$ and $B = C = - 1$

And $\frac{10}{\left(x - 1\right) \left({x}^{2} + 9\right)} = \frac{1}{x - 1} + \frac{- x - 1}{{x}^{2} + 9} = \frac{1}{x - 1} - \frac{x}{{x}^{2} + 9} - \frac{1}{{x}^{2} + 9}$,

so

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

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