Question

Find: $\int \frac{e^x(x-3)}{(x-1{)}^3}dx$

Answer 1

$\int \frac{(x-3)}{(x-1)}{e}^x.dx.$

$=\int \frac{x-1-2}{(x-1{)}^3}.e^x.dx.$

$=\int e^x(\frac{x-1}{(x-1)3}-\frac{2}{(x-1{)}^3})dx$

$=\int e^x(\frac{1}{(x-1{)}^2}-\frac{2}{(x-1{)}^3})dx$

put $f\left(x\right)=\frac{1}{(x-1{)}^2}\to f^{\prime }\left(x\right)=\frac{-2}{(x-1{)}^3}$

Thus

$\int e^x(\frac{1}{(x-1{)}^2}-\frac{2}{(x-1{)}^3})dx=\frac{e^x}{(x-1{)}^2}+C$