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

Apr 15, 2018

$4 \ln | \frac{x - 1}{x} | + \frac{3}{x} + C , \mathmr{and} , \ln {\left(1 - \frac{1}{x}\right)}^{4} + \frac{3}{x} + C$.

#### Explanation:

Here is a way to find the Integral $I = \int \frac{x + 3}{{x}^{2} \left(x - 1\right)} \mathrm{dx}$,

without using partial fractions.

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

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

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

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

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

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

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

$= 4 \int \left\{\frac{1}{x - 1} - \frac{1}{x}\right\} \mathrm{dx} + \frac{3}{x}$,

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

$\Rightarrow I = 4 \ln | \frac{x - 1}{x} | + \frac{3}{x} + C , \mathmr{and} , \ln {\left(1 - \frac{1}{x}\right)}^{4} + \frac{3}{x} + C$.

Enjoy Maths.!