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

Jul 3, 2016

$I = x - \frac{2}{3} \ln \left(x - 1\right) - \frac{10}{3} \ln \left(x + 2\right) + C ,$ OR
$I = x - \frac{2}{3} \ln \left(x - 1\right) {\left(x + 2\right)}^{5} + C ,$ OR
$I = x - \ln {\left(x - 1\right)}^{\frac{2}{3}} {\left(x + 2\right)}^{\frac{10}{3}} + C .$

#### Explanation:

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

The Degree of Poly. in $N r .$ = $2$ = that of poly in $D r .$

Hence, it is Improper Rational fun. To make it Proper, usually Long Division is performrd, but, here we proceed as under :

We write, $N r . = {x}^{2} - 3 x = {x}^{2} + x - 2 \left[i . e . , = D r\right] - 4 x + 2$, so

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

Hence, $I = \int \left[1 - \frac{4 x - 2}{\left(x - 1\right) \left(x + 2\right)}\right] \mathrm{dx} = \int 1 \mathrm{dx} - {I}_{1} = x - {I}_{1} ,$

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

To evaluate ${I}_{1}$, we have to split $\frac{4 x - 2}{\left(x - 1\right) \left(x + 2\right)} \ldots \ldots \left(1\right)$ using Partial Fractions as A/(x-1)+B/(x+2)={A(x+2)+B(x-1)}/{(x-1)(x+2)}.....(2); A,B in RR.

Since $\left(1\right) \mathmr{and} \left(2\right)$ are equal, we get $: A \left(x + 2\right) + B \left(x - 1\right) = 4 x - 2 , \forall x .$

In particular, $x = 1 \Rightarrow 3 A = 2 \Rightarrow A = \frac{2}{3}$ and $x = - 2 \Rightarrow B = \frac{10}{3.}$

Accordingly, ${I}_{1} = \int \frac{\frac{2}{3}}{x - 1} \mathrm{dx} + \int \frac{\frac{10}{3}}{x + 2} \mathrm{dx}$
$= \frac{2}{3} \ln | x - 1 | + \frac{10}{3} \ln | x + 2 | .$ Finally, we altogether get,

$I = x - \frac{2}{3} \ln \left(x - 1\right) - \frac{10}{3} \ln \left(x + 2\right) + C$, or, using the Rules of Log,

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