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

Nov 22, 2016

The answer is $= \ln \left(x + 1\right) - \frac{1}{2} \ln \left({x}^{2} + 2 x + 2\right) + C$

#### Explanation:

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

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

So,
$1 = A \left({x}^{2} + 2 x + 2\right) + \left(B x + C\right) \left(x + 1\right)$

Let $x = 0$, $\implies$, $1 = 2 A + C$

Let $x = - 1$, $\implies$, $1 = A$

coefficients of x^2, $0 = A + B$ $\implies$, $B = - 1$

#C=-1

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

$= \frac{1}{x + 1} - \frac{x + 1}{{x}^{2} + 2 x + 2}$

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

$\int \frac{\mathrm{dx}}{x + 1} = \ln \left(x + 1\right)$

For the second integral, we use a substitution

$u = {x}^{2} + 2 x + 2$

$\mathrm{du} = \left(2 x + 2\right) \mathrm{dx} = 2 \left(x + 1\right) \mathrm{dx}$

$\int \frac{\left(\left(x + 1\right)\right) \mathrm{dx}}{{x}^{2} + 2 x + 2} = \frac{1}{2} \int \frac{\mathrm{du}}{u} = \frac{1}{2} \ln u$

$= \frac{1}{2} \ln \left({x}^{2} + 2 x + 2\right)$

Putting it all together,

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