Question

If $\underset{0}{\overset{1}{\int }}\frac{e^x}{1+x}dx=k,$ then $\underset{0}{\overset{1}{\int }}\frac{e^x}{(1+x{)}^2}dx$ is equal to

• A. $k+\frac{e}{2}+1$
• B. $k-\frac{e}{2}+1$
• C. $k-\frac{e}{2}$
• D. $k+\frac{e}{2}$

Answer 1

The correct option is B $k-\frac{e}{2}+1$

$k=\underset{0}{\overset{1}{\int }}\frac{e^x}{1+x}dx$
Applying integration by parts
$k=\underset{0}{\overset{1}{\int }}\underset{I}{\frac{1}{1+x}}\cdot {\underset{II}{e}}^{x}dx\Rightarrow k={\left[\frac{e^x}{1+x}\right]}_0^1-\underset{0}{\overset{1}{\int }}\frac{-e^x}{(1+x{)}^2}\Rightarrow k=(\frac{e^1}{2}-e^0)+\underset{0}{\overset{1}{\int }}\frac{e^x}{(1+x{)}^2}\Rightarrow \underset{0}{\overset{1}{\int }}\frac{e^x}{(1+x{)}^2}=k-\frac{e}{2}+1$