Integrals — Class 12 Maths Solution

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

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

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