Question
Evaluate the following integrals:$\int\text{e}^{\text{x}}\frac{1+\text{x}}{(2+\text{x})^2}\text{dx}$

Answer

Let $\text{I}=\int\text{e}^{\text{x}}\frac{1+\text{x}}{(2+\text{x})^2}\text{dx}$
$=\int\text{e}^{\text{x}}\Big(\frac{\text{x}+2-1}{(2+\text{x})^2}\Big)\text{dx}$
$=\int\text{e}^{\text{x}}\bigg\{\frac{1}{\text{x}+2}-\frac{1}{(\text{x}+2)^2}\bigg\}\text{dx}$
$=\int\text{e}^{\text{x}}\frac{1}{\text{x}+2}\text{dx}-\int\text{e}^{\text{x}}\frac{1}{(\text{x}+2)^2}\text{dx}$
Integrating by parts
$=\text{e}^{\text{x}}\frac{1}{\text{x}+2}-\int\text{e}^{\text{x}}\Big(\frac{\text{d}}{\text{dx}}\Big(\frac{1}{\text{x}+2}\Big)\Big)\text{dx}-\int\text{e}^{\text{x}}\frac{1}{(\text{x}+2)^2}\text{dx}$
$=\text{e}^{\text{x}}\frac{1}{\text{x}+2}+\int\text{e}^{\text{x}}\frac{1}{(\text{x}+2)^2}\text{dx}-\int\text{e}^{\text{x}}\frac{1}{(\text{x}+2)^2}\text{dx}$
$=\frac{\text{e}^{\text{x}}}{\text{x}+2}+\text{C}$

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