# How do you integrate int (3x^2 + 10x -5)/ ( (x+1)^2(x-2) ) using partial fractions?

Jul 19, 2017

$\int \frac{3 {x}^{2} + 10 x - 5}{{\left(x + 1\right)}^{2} \left(x - 2\right)} \mathrm{dx} = - \frac{4}{x + 1} + 3 \ln | x - 2 | + C$

#### Explanation:

$\frac{3 {x}^{2} + 10 x - 5}{{\left(x + 1\right)}^{2} \left(x - 2\right)} = \frac{A}{x + 1} + \frac{B}{x + 1} ^ 2 + \frac{C}{x - 2} =$

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

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

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

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

So we know now :

$A + C = 3$
$- A + B + 2 C = 10$
$- 2 A - 2 B + C = - 5$

From this system we get :

$A = 0 , B = 4 , C = 3$

Let's get to the integral now:

$\int \frac{3 {x}^{2} + 10 x - 5}{{\left(x + 1\right)}^{2} \left(x - 2\right)} \mathrm{dx} = \int \left(\frac{4}{x + 1} ^ 2 + \frac{3}{x - 2}\right) \mathrm{dx} =$

$\int \frac{4}{x + 1} ^ 2 \mathrm{dx} + \int \frac{3}{x - 2} \mathrm{dx} =$

$\int 4 {\left(x + 1\right)}^{- 2} \mathrm{dx} + 3 \ln | x - 2 | =$

$- 4 {\left(x + 1\right)}^{- 1} + 3 \ln | x - 2 | =$

$- \frac{4}{x + 1} + 3 \ln | x - 2 | + C$