Question

${\int }_0^1\frac{dx}{e^x+e^{-x}}=$

• A. $ta{n}^{-1}e$
• B. $\frac{\pi }{4}$
• C. $ta{n}^{-1}e-\frac{\pi }{4}$
• D. $ta{n}^{-1}e+\frac{\pi }{4}$

Answer 1

Answer: (C)

${\int }_0^1\frac{e^x}{1+e^{2x}}dx={\int }_0^1\frac{d{e}^x}{1+e^{2x}}$
$=ta{n}^{-1}\left(e^x\right){\int }_0^1$
$=ta{n}^{-1}\left(e\right)-ta{n}^{-1}\left(e^{\circ }\right)$
$=ta{n}^{-1}\left(e\right)-ta{n}^{-1}\left(1\right)$
$=ta{n}^{-1}\left(e\right)-\frac{\pi }{4}$
So, ${\int }_0^1\frac{e^x}{1+e^{2x}}dx=ta{n}^{-1}e-\frac{\pi }{4}$