If \(\int_{0}^{1}\displaystyle\frac{e^x}{1+x}\,dx = m\), then the value of \(\int_{0}^{1}\displaystyle\frac{e^x}{(1+x)^2}\,dx\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(m - \displaystyle\frac{e}{2} + 1\)
By integration by parts on m, taking \(\frac{1}{1+x}\) as the first function and \(e^x\) as the second:
\(\displaystyle m = \int_{0}^{1} \frac{1}{1+x}\,e^x\,dx\)
\(\displaystyle = \left[\frac{e^x}{1+x}\right]_{0}^{1} - \int_{0}^{1} \left(-\frac{1}{(1+x)^2}\right)e^x\,dx\)
\(\displaystyle = \left(\frac{e}{2} - 1\right) + \int_{0}^{1} \frac{e^x}{(1+x)^2}\,dx\)
So
\(\displaystyle \int_{0}^{1} \frac{e^x}{(1+x)^2}\,dx = m - \frac{e}{2} + 1\)