Question:
Evaluate: $\int \frac{2 \mathrm{x}+3}{(\mathrm{x}-1)^{2}} \mathrm{dx}$
Solution:
The above equation can be written as
$\Rightarrow \int \frac{2 x-2+2+3}{(x-1)^{2}}$
$\Rightarrow \int \frac{2(x-1)+5}{(x-1)^{2}}$
$\Rightarrow 2 \int \frac{1 . d x}{(x-1)}+5 \int \frac{1 . d x}{(x-1)^{2}}$
We know $\int \mathrm{x} \mathrm{dx}=\frac{x^{\mathrm{n}}}{\mathrm{n}+1} ; \int \frac{1}{\mathrm{x}} \mathrm{dx}=\ln \mathrm{x}$
$\Rightarrow 2 \ln (\mathrm{x}-1)+5 \int(\mathrm{x}-1)^{-2} \mathrm{dx}$
$\Rightarrow 2 \ln (\mathrm{x}-1)+5 \int \frac{(\mathrm{x}-1)^{-1}}{-1} \mathrm{dx}$
$\Rightarrow 2 \ln (\mathrm{x}-1)-\frac{5}{(\mathrm{x}-1)}+\mathrm{c} .$ (Where $\mathrm{c}$ is an arbitrary constant)