Question

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

• A. ${\tan }^{-1}e-\frac{\pi }{4}$
• B. ${\tan }^{-1}e+\frac{\pi }{4}$
  
• C. $\tan e-\frac{\pi }{4}$
  
• D. None of these

Answer 1

The correct option is A

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

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

Let $e^x=t⟹e^{x}dx=dt$

$x\to 0\to 1$

$⟹t\to 1\to e$

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

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

$⟹{\tan }^{-1}e-\frac{\pi }{4}$