Integrals — Class 12 Maths Solution

Let$I = \int {\cfrac{{\left( {x - 3} \right){e^x}}}{{{{\left( {x - 1} \right)}^3}}}dx} = \int {\cfrac{{{e^x}\left( {\left( {x - 1} \right) - 2} \right)}}{{{{\left( {x - 1} \right)}^3}}}} dx$

$= \int {{e^x}\left[ {\cfrac{1}{{{{\left( {x - 1} \right)}^2}}} - \cfrac{2}{{{{\left( {x - 1} \right)}^3}}}} \right]dx}$

$= \int {{e^x}\left[ {\cfrac{1}{{{{\left( {x - 1} \right)}^2}}} + \cfrac{d}{{dx}}\left( {\cfrac{1}{{{{\left( {x - 1} \right)}^2}}}} \right)} \right]dx}$

$= \cfrac{{{e^x}}}{{{{\left( {x - 1} \right)}^2}}} + C$