The value of the definite integral \(\int_{0}^{1}\displaystyle e^x\frac{(1-x)^2}{(1+x^2)^2}\,dx\) is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{e}{2} - 1\)
The integrand has the form \(e^x[f(x) + f'(x)]\), and by the standard result \(\int e^x[f(x) + f'(x)]\,dx = e^x f(x) + C\). Let
\( \displaystyle f(x) = \frac{1}{1 + x^2} \)
\( \displaystyle f'(x) = -\frac{2x}{(1 + x^2)^2} \)
Then
\( \displaystyle f(x) + f'(x) = \frac{1 + x^2 - 2x}{(1 + x^2)^2} \)
\( \displaystyle = \frac{(1 - x)^2}{(1 + x^2)^2} \)
which is exactly the fraction in the integral. So
\( \displaystyle \int_{0}^{1} e^x\frac{(1-x)^2}{(1+x^2)^2}\,dx = \left[\frac{e^x}{1 + x^2}\right]_{0}^{1} \)
\( \displaystyle = \frac{e}{2} - \frac{1}{1} \)
\( \displaystyle = \frac{e}{2} - 1 \)