Question

If $\underset{0}{\overset{1}{\int }}\frac{e^t}{1+t}dt=a,$ then $\underset{0}{\overset{1}{\int }}\frac{e^t}{(1+t{)}^2}$ is equal to

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

Answer 1

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

We have,
$\underset{0}{\overset{1}{\int }}\frac{e^t}{1+t}dt=a$
Using integration by parts, we get
$\Rightarrow a={\left(\frac{e^t}{1+t}\right)}_0^1+\underset{0}{\overset{1}{\int }}\frac{e^t}{(1+t{)}^2}dt\Rightarrow a=\frac{e}{2}-1+\underset{0}{\overset{1}{\int }}\frac{e^t}{(1+t{)}^2}dt\Rightarrow \underset{0}{\overset{1}{\int }}\frac{e^t}{(1+t{)}^2}dt=a+1-\frac{e}{2}$