# How do you use partial fraction decomposition to decompose the fraction to integrate 4/((x^2+9)(x+1))?

May 19, 2018

int4/((x²+9)(x+1))dx=-1/5ln(x²+9)+2/15arctan(x/3)+2/5ln(|x+1|)+ C
$C \in \mathbb{R}$

#### Explanation:

int4/((x²+9)(x+1))dx=4int1/((x²+9)(x+1))dx
let 1/((x²+9)(x+1))=(Ax+B)/(x²+9)+C/(x+1)
1=(Ax+B)(x+1)+C(x²+9)
1=Ax²+Ax+Bx+B+Cx²+9C

1=(A+C)x²+(A+B)x+9C
By identification :

$A + C = 0$
$A + B = 0$
$9 C + B = 1$

$A + C = 0$
$B - C = 0$
$9 C + B = 1$

$A + C = 0$
$B - C = 0$
$10 C = 1$

$A = - \frac{1}{10}$
$B = \frac{1}{10}$
$C = \frac{1}{10}$

So: 4int1/((x²+9)(x+1))dx=4int((-1/10x+1/10)/(x²+9)+(1/10)/(x+1))dx

=-4/10intx/(x²+9)dx+4/10int1/(x²+9)dx+4/10int1/(x+1)dx
=-2/10int(2x)/(x²+9)dx+2/5int1/(x²+3²)dx+2/5int1/(x+1)dx
Because $\int \frac{u '}{u} \mathrm{du} = \ln \left(| u |\right)$, int1/(x²+b²)dx=1/b*arctan(x/b),
we have :

int4/((x²+9)(x+1))dx=-1/5ln(x²+9)+2/15arctan(x/3)+2/5ln(|x+1|)+ C
$C \in \mathbb{R}$

May 19, 2018

See the explanation below

#### Explanation:

The decomposition into partial fractions is

$\frac{4}{\left({x}^{2} + 9\right) \left(x + 1\right)} = \frac{A x + B}{{x}^{2} + 9} + \frac{C}{x + 1}$

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

The denominators are the same, compare the numerators

$4 = \left(A x + B\right) \left(x + 1\right) + C \left({x}^{2} + 9\right)$

Let $x = - 1$, $\implies$, $4 = 10 C$, $\implies$, $C = \frac{2}{5}$

The coefficients of ${x}^{2}$ are

$0 = A + C$, $\implies$, $A = - C = - \frac{2}{5}$

Let $x = 0$

$4 = B + 9 C$, $\implies$, $B = 4 - 9 C = 4 - \frac{18}{5} = \frac{2}{5}$

Therefore,

$\frac{4}{\left({x}^{2} + 9\right) \left(x + 1\right)} = \frac{- \frac{2}{5} x + \frac{2}{5}}{{x}^{2} + 9} + \frac{\frac{2}{5}}{x + 1}$

So, the integral is

$I = \int \frac{4 \mathrm{dx}}{\left({x}^{2} + 9\right) \left(x + 1\right)} = \int \frac{\left(- \frac{2}{5} x + \frac{2}{5}\right) \mathrm{dx}}{{x}^{2} + 9} + \int \frac{\frac{2}{5} \mathrm{dx}}{x + 1}$

$= \frac{2}{5} \ln \left(| x + 1 |\right) + {I}_{1}$

${I}_{1} = \frac{2}{5} \int \frac{\left(- x + 1\right) \mathrm{dx}}{{x}^{2} + 9}$

$= \frac{2}{5} \int \frac{- x \mathrm{dx}}{{x}^{2} + 9} + \frac{2}{5} \int \frac{\mathrm{dx}}{{x}^{2} + 9}$

$= - \frac{2}{5} \cdot \frac{1}{2} \ln \left({x}^{2} + 9\right) + \frac{2}{5} \cdot \frac{1}{9} \cdot 3 \arctan \left(\frac{x}{3}\right)$

Finally,

$\int \frac{4 \mathrm{dx}}{\left({x}^{2} + 9\right) \left(x + 1\right)} = \frac{2}{5} \ln \left(| x + 1 |\right) - \frac{1}{5} \ln \left({x}^{2} + 9\right) + \frac{2}{15} \arctan \left(\frac{x}{3}\right) + C$