WBJEE · Maths · Indefinite Integration
The value of the integral \(\int \frac{d x}{\left(e^x+e^{-x}\right)^2}\) is
- A \(\frac{1}{2}\left(\mathrm{e}^{2 \mathrm{x}}+1\right)+\mathrm{C}\)
- B \(\frac{1}{2}\left(e^{-2 x}+1\right)+C\)
- C \(-\frac{1}{2}\left(e^{2 x}+1\right)^{-1}+C\)
- D \(\frac{1}{4}\left(e^{2 x}-1\right)+C\)
Answer & Solution
Correct Answer
(C) \(-\frac{1}{2}\left(e^{2 x}+1\right)^{-1}+C\)
Step-by-step Solution
Detailed explanation
Hints: \(\int \frac{e^{2 x} d x}{\left(e^{2 x}+1\right)^2} \quad e^x=t ; e^x d x=d t\) \[ =\frac{1}{2} \int \frac{2 \mathrm{tdt}}{\left(\mathrm{t}^2+1\right)^2}=\frac{1}{2}\left\{-\frac{1}{\left(t^2+1\right)}\right\}+c=-\frac{1}{2\left(\mathrm{e}^{2 x}+1\right)}+c \]
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