Question

$\int_{}^{}e^{x}\left( \tan^{- 1}x + \frac{2x}{(1 + x^{2})^{2}} \right)$.dx is equal to

• A. e^x$\left( \cot^{- 1}x - \frac{1}{1 + x^{2}} \right)$ + c
• B. e^x$\left( \tan^{- 1}x + \frac{1}{1 + x^{2}} \right)$ + c
• C. e^x$\left( \tan^{- 1}x - \frac{1}{1 + x^{2}} \right)$ + c
• D. e^x tan^–1 x + c

Answer 1

Answer: (C)

e^x$\left( \tan^{- 1}x - \frac{1}{1 + x^{2}} \right)$ + c

Answer 2

$\left( \tan^{- 1}x + \frac{2x}{(1 + x^{2})^{2}} \right)$dx

= $\left( \tan^{- 1}x + \frac{1}{1 + x^{2}} \right)$ $$\downarrow \downarrow $$ $f(x)f'(x)$ dx –

$\left( \frac{1}{1 + x^{2}} - \frac{2x}{(1 + x^{2})^{2}} \right)$ $${\downarrow \downarrow }{g(x)g'(x)}$$

= e^x tan^–1(x) – e^x $\frac{1}{1 + x^{2}}$ + c

= e^x $\left( \tan^{- 1}x - \frac{1}{1 + x^{2}} \right)$ + c