Question

$\underset{0}{\overset{1}{\int }}\frac{e^x}{1+e^{2x}}dx=$

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

Answer 1

Answer: (D)

${\tan }^{-1}e-\frac{\pi }{4}$

Given ${\int }_0^1\frac{e^x}{1+e^{2x}}dx$

Let $e^x=t⟹e^{x}dx=dt$

$xto0\to 1$

$⟹t\to 1\to e$

${\int }_1^e\frac{1}{1+t^2}dt$

${ta{n}^{-1}x|}_1^e$

$⟹{\tan }^{-1}e-\frac{\pi }{4}$