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Q.
The value of the definite integral $\int\limits_{-1}^1 \frac{ dx }{\left(1+ e ^{ x }\right)\left(1+ x ^2\right)}$ is
Integrals
Solution:
$ I=\int\limits_{-1}^1 \frac{d x}{\left(1+e^x\right)\left(1+x^2\right)}$ ....(1)
$= \int\limits_{-1}^1 \frac{d x}{1+e^{-x}} \cdot \frac{1}{1+x^2} $(using King)
$I =\int\limits_{-1}^1 \frac{e^x d x}{\left(1+e^x\right)\left(1+x^2\right)}$....(2)
adding (1) and (2)
$2 I=\int\limits_{-1}^1 \frac{\left(1+e^x\right) d x}{\left(1+e^x\right)\left(1+x^2\right)}=\int\limits_{-1}^1 \frac{d x}{\left(1+x^2\right)}=2 \int\limits_0^1 \frac{d x}{\left(1+x^2\right)} $
$I=\int\limits_0^1 \frac{d x}{\left(1+x^2\right)}=\tan ^{-1}(1)=\pi / 4$
[convert it into value of definite integral ' $T$ ' is same as ]