Question

$\int_{}^{}e^{\tan^{- 1}x}$ (1 + x + x²)d(cot^–1 x) is equal to-

• A. $- e^{\tan^{- 1}x} + C$
• B. $e^{\tan^{- 1}x} + C$
• C. $- xe^{\tan^{- 1}x} + C$
• D. x $e^{\tan^{- 1}x} + C$

Answer 1

Answer: (C)

$- xe^{\tan^{- 1}x} + C$

Answer 2

I = –

Let tan^–1x = t

So I = –$\int \mathrm { e } ^ { \mathrm { t } } \left( 1 + \tan \mathrm { t } + \tan ^ { 2 } \mathrm { t } \right) \mathrm { dt }$ = –

= – e^t . tan t + C = – x