Question

If ${\int }_0^1\frac{e^x}{1+t}dt=a$ then ${\int }_0^1\frac{e^x}{{(1+t)}^2}dt=$

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

Answer 1

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

$\begin{array}{c}Givena={\int }_0^1\frac{e^t}{1+t}dt\\ now{\int }_0^1\frac{e^t}{{\left(1+t\right)}^2}=a+1-\frac{e}{2}\\ nowintegratinga={\left(\frac{1}{1+t}{e}^t\right)}_0^1+{\int }_0^1\frac{e^t}{{\left(1+t\right)}^2}dt\left[integratingbyparts\right]\\ a=\frac{e^1}{2}-\frac{e^0}{1}+{\int }_0^1\frac{e^t}{{\left(1+t\right)}^2}dt\\ a=\frac{e}{2}-1+{\int }_0^1\frac{e^t}{{\left(1+t\right)}^2}\end{array}$