# How do you find the indefinite integral of int (x(x+2))/(x^3+3x^2-4)?

Jun 1, 2018

$\frac{1}{3} \ln | {\left(x + 2\right)}^{2} \left(x - 1\right) | + C , O R , \ln | {\left(x + 2\right)}^{\frac{2}{3}} {\left(x - 1\right)}^{\frac{1}{3}} | + C$.

#### Explanation:

Let, $I = \int \frac{x \left(x + 2\right)}{{x}^{3} + 3 {x}^{2} - 4} \mathrm{dx}$.

The sum of the co-effs. of the poly. in the Dr. is $0$.

$\therefore \left(x - 1\right)$ is its factor.

We have, ${x}^{3} + 3 {x}^{2} - 4 = \underline{{x}^{3} - {x}^{2}} + \underline{4 {x}^{2} - 4 x} + \underline{4 x - 4}$,

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

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

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

$\therefore \frac{x \left(x + 2\right)}{{x}^{3} + 3 {x}^{2} - 4} = \frac{x \left(x + 2\right)}{\left(x - 1\right) {\left(x + 2\right)}^{2}} = \frac{x}{\left(x - 1\right) \left(x + 2\right)}$.

Observe that, $2 \left(x - 1\right) + 1 \left(x + 2\right) = 3 x$. Therefore,

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

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

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

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

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

$= \frac{1}{3} \int \left\{\frac{2}{x + 2} + \frac{1}{x - 1}\right\} \mathrm{dx}$,

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

$\therefore I = \frac{1}{3} \ln | {\left(x + 2\right)}^{2} \left(x - 1\right) | + C , \mathmr{and} ,$

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

Enjoy Maths.!