Question

$\text{∫}\frac{\mathrm{x}{\mathrm{e}}^{\mathrm{x}}}{{(1+\mathrm{x})}^2}\mathrm{is}\mathrm{equal}\mathrm{to}:$

Answer 1

Find the integral of the given expression

Given: $\mathrm{I}=\text{∫}\frac{\mathrm{x}{\mathrm{e}}^{\mathrm{x}}}{{(1+\mathrm{x})}^2}\text{dx}$

$\mathrm{I}=\text{∫}\frac{(\mathrm{x}+1-1){\mathrm{e}}^{\mathrm{x}}}{{(1+\mathrm{x})}^2}\text{dx}$

$\mathrm{I}=\text{∫}\frac{{\mathrm{e}}^{\mathrm{x}}}{(1+\mathrm{x})}\text{dx}-\text{∫}\frac{{\mathrm{e}}^{\mathrm{x}}}{{(1+\mathrm{x})}^2}\text{dx}$

Now using the integration by parts method in the first integral, we get;

$$\begin{gathered} \mathrm{I}=\frac{e^x}{(1+x)}+\text{∫}\left(\frac{{\mathrm{e}}^{\mathrm{x}}}{{(1+\mathrm{x})}^2}\right)\text{dx}-\text{∫}\left(\frac{{\mathrm{e}}^{\mathrm{x}}}{{(1+\mathrm{x})}^2}\right)\text{dx + C} \\ \mathrm{I}=\frac{e^x}{(1+x)}+\text{∫}\left(\frac{1}{{(1+\mathrm{x})}^2}\right){\text{e}}^x\text{dx}-\text{∫}\left(\frac{{\mathrm{e}}^{\mathrm{x}}}{{(1+\mathrm{x})}^2}\right)\text{dx + C} \\ \mathrm{I}=\frac{e^x}{(1+x)}+C \end{gathered}$$

Hence, the answer is $\frac{{\mathrm{e}}^{\mathrm{x}}}{(1+\mathrm{x})}+\mathrm{C}$, where $C$ is an arbitrary constant.