∫ e t a n − 1 x ( 1 + x 2 ) [ ( s e c − 1 √ 1 + x 2 ) 2 + c o s − 1 ( 1 − x 2 1 + x 2 ) ] d x ( x > 0 ) Options: A . &nbsp e t a n − 1 x . t a n − 1 x + C B . &nbsp e t a n − 1 x . ( t a n − 1 x ) 2 2 + C C . &nbsp e t a n − 1 x . ( s e c − 1 ( √ 1 + x 2 ) ) 2 + C D . &nbsp e t a n − 1 x . ( c o s − 1 ( √ 1 + x 2 ) ) 2 + C

Answer: Option C : C note that s e c − 1 √ 1 + x 2 = t a n − 1 x ; c o s − 1 ( 1 − x 2 1 + x 2 ) = 2 t a n − 1 x for x > 0 I = ∫ e t a n − 1 x 1 + x 2 ( ( t a n − 1 x ) 2 + 2 t a n − 1 x ) d x Put t a n − 1 x = t = ∫ e t ( t 2 + 2 t ) d t = e t . t 2 = e t a n − 1 x ( ( t a n − 1 x ) 2 ) + C

∫etan−1x(1+x2)[(sec−1√1+x2)2+cos−1(1−

Discussion Forum : Indefinite Integration

Question -

$\int \frac{e^{t a n^{- 1} x}}{(1 + x^{2})} [{(s e c^{- 1} \sqrt{1 + x^{2}})}^{2} + c o s^{- 1} (\frac{1 - x^{2}}{1 + x^{2}})] d x (x > 0)$

Options:

A .

e^{t a n^{- 1} x} . t a n^{- 1} x + C

B .

\frac{e^{t a n^{- 1} x} . (t a n^{- 1} x)^{2}}{2} + C

C .

e^{t a n^{- 1} x} . {(s e c^{- 1} (\sqrt{1 + x^{2}}))}^{2} + C

D .

e^{t a n^{- 1} x} . {(c o s^{- 1} (\sqrt{1 + x^{2}}))}^{2} + C

Answer: Option C
:
C
note that

s e c^{- 1} \sqrt{1 + x^{2}} = t a n^{- 1} x; c o s^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) = 2 t a n^{- 1} x

for x > 0

I = \int \frac{e^{t a n^{- 1 x}}}{1 + x^{2}} ((t a n^{- 1} x)^{2} + 2 t a n^{- 1} x) d x

Put

t a n^{- 1} x = t

= \int e^{t} (t^{2} + 2 t) d t = e^{t} . t^{2} = e^{t a n^{- 1} x} ((t a n^{- 1} x)^{2}) + C

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Discussion Forum : Indefinite Integration

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