Question
Evaluate the following integrals:
$\int\text{e}^{\text{x}}\frac{\text{x}-1}{(\text{x}+1)^3}\text{dx}$

Answer

Let $\text{I}=\int\text{e}^{\text{x}}\frac{\text{x}+1-2}{(\text{x}+1)^3}\text{dx}$
$=\int\text{e}^{\text{x}}\Big\{\frac{1}{(\text{x}+1)^2}+\frac{-2}{({\text{x}+1})^3}\Big\}\text{dx}$
$=\int\text{e}^{\text{x}}\frac{1}{(\text{x}+1)^2}\text{dx}+\int\text{e}^{\text{x}}\frac{(-2)}{(\text{x}+1)^3}\text{dx}$
Integrating by parts
$=\text{e}^{\text{x}}\frac{1}{(\text{x}+1)^2}-\int\text{e}^{\text{x}}\Big(\frac{\text{d}}{\text{dx}}(\text{x}+1)^{-2}\Big)\text{dx}+\int\text{e}^{\text{x}}\frac{(-2)}{(\text{x}+1)^3}\text{dx}$
$=\text{e}^{\text{x}}\frac{1}{(\text{x}+1)^2}-\int\text{e}^{\text{x}}\frac{(-2)}{\text{(x+1)}^3}\text{dx}+\int\text{e}^{\text{x}}\frac{(-2)}{(\text{x}+1)^3}\text{dx}$
$=\text{e}^{\text{x}}\frac{1}{(\text{x}+1)}+\text{C}$

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