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

Jul 18, 2016

$\frac{1}{x} + \ln | {x}^{2} {\left(x + 2\right)}^{3} | + {C}_{1}$

#### Explanation:

The $D r . = {x}^{3} + 2 {x}^{2} = {x}^{2} \left(x + 2\right)$

So, we are having linear repeated factors. In such case, we take,

$\frac{5 {x}^{2} + 3 x - 2}{{x}^{3} + 2 {x}^{2}} = \frac{A}{x} + \frac{B}{x} ^ 2 + \frac{C}{x + 2} \ldots . \left(1\right) , w h e r e , A , B , C \in \mathbb{R}$

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

$\Rightarrow 5 {x}^{2} + 3 x - 2 = A x \left(x + 2\right) + B \left(x + 2\right) + C {x}^{2.} \ldots \ldots \ldots \ldots . \left(2\right)$

Eventhough $\left(2\right)$ is true $\forall x \in \mathbb{R}$, we select,

$x = 0 \Rightarrow - 2 = 2 B \Rightarrow B = - 1$

$x = - 2 \Rightarrow 20 - 6 - 2 = 12 = 4 C \Rightarrow C = 3$

$x = 1 \Rightarrow 5 + 3 - 2 = 6 = 3 A + 3 B + C = 3 A - 3 + 3 = 3 A$

$\Rightarrow A = 2$

$\therefore I = \int \frac{5 {x}^{2} + 3 x - 2}{{x}^{3} + 2 {x}^{2}} \mathrm{dx}$

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

$= 2 \ln x + \frac{1}{x} + 3 \ln | x + 2 |$

$= \frac{1}{x} + \ln | {x}^{2} {\left(x + 2\right)}^{3} | + {C}_{1}$

Hope, this will be of a little Help! Enjoy Maths. & spread the Joy!