Question

If ${\int }_0^1\frac{e^{t}dt}{1+t}=A$ then the value of ${\int }_0^1\frac{e^{t}dt}{(1+t{)}^2}$ is-

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

Answer 1

Answer: (D)

$A-\frac{e}{2}+1$

${\int }_0^1\frac{e^{t}dt}{1+t}=A$

${\left|\frac{e^t}{1+t}\right|}_0^1-{\int }_0^1\frac{-1\times e^t}{(1+t{)}^2}dt=A$ (by parts)

$\frac{e}{2}-1+{\int }_0^1\frac{e^{t}dt}{(1+t^2)}=A$

$\Rightarrow {\int }_0^1\frac{e^t}{(1+t{)}^2}dt=A+1-\frac{e}{2}$