Question:
Evaluate: $\int \frac{2 x-1}{(x-1)^{2}} d x$
Solution:
In this question degree of denominator is larger than that of numerator so we need to manipulate numerator.
$\Rightarrow \int \frac{2 x+2-2-1}{(x-1)^{2}}$
$\Rightarrow \int \frac{2(x-1)-1}{(x-1)^{2}}$
$\Rightarrow \int \frac{2}{x-1} d x-\frac{1}{(x-1)^{2}} d x$
We know $\int x d x=\frac{x^{n}}{n+1} ; \int \frac{1}{x} d x=\ln x$
$\Rightarrow 2 \ln (x-1)-\int(x-1)^{-2} d x$
$\Rightarrow 2 \ln (\mathrm{x}-1)-\frac{1}{\mathrm{x}-1}+\mathrm{c}$. (where $\mathrm{c}$ is some arbitrary constant)