# Partial fractions \int5/((x^2+2x+2)(x-1))?

## I am getting results like this, are they ok so far? $A + C = 0$, $B - A + 2 C = 0$, $2 C - B = 5$

Apr 13, 2018

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

#### Explanation:

We need

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

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

Perform the decomposition into partial fractions

$\frac{5}{\left(x - 1\right) \left({x}^{2} + 2 x + 2\right)} = \frac{A}{x - 1} + \frac{B x + C}{{x}^{2} + 2 x + 2}$

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

The denominatoras are the same, compare the numerators

$5 = A \left({x}^{2} + 2 x + 2\right) + \left(B x + C\right) \left(x - 1\right)$

Let $x = 1$, $\implies$, $5 = 5 A$, $\implies$, $A = 1$

Let $x = 0$, $\implies$, $5 = 2 A - C$, $\implies$, $C = 2 A - 5 = 2 - 5 = - 3$

Coefficients of ${x}^{2}$,

$0 = A + B$, $\implies$, $B = - A = - 1$

Therefore,

$\frac{5}{\left(x - 1\right) \left({x}^{2} + 2 x + 2\right)} = \frac{1}{x - 1} + \frac{- x - 3}{{x}^{2} + 2 x + 2}$

So,

$\int \frac{5 \mathrm{dx}}{\left(x - 1\right) \left({x}^{2} + 2 x + 2\right)} = \int \frac{1 \mathrm{dx}}{x - 1} - \int \frac{\left(x + 3\right) \mathrm{dx}}{{x}^{2} + 2 x + 2}$

$= \ln \left(| x - 1 |\right) - I$

And

$I = \int \frac{\left(x + 3\right) \mathrm{dx}}{{x}^{2} + 2 x + 2} = \int \frac{\left(\frac{1}{2} \left(2 x + 2\right) + 2\right) \mathrm{dx}}{{x}^{2} + 2 x + 2}$

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

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

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

$= \arctan \left(x + 1\right)$

Finally,

$\int \frac{5 \mathrm{dx}}{\left(x - 1\right) \left({x}^{2} + 2 x + 2\right)} = \ln \left(| x - 1 |\right) - \frac{1}{2} \ln \left(| {x}^{2} + 2 x + 2 |\right) - 2 \arctan \left(x + 1\right) + C$