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Question: \[\int_{}^{}\sqrt{\frac{e^{x} + a}{e^{x} - a}}dx =\]...

∫ex+aex−adx=\int_{}^{}\sqrt{\frac{e^{x} + a}{e^{x} - a}}dx =

A

cos⁡h−1(ex/a)+sec⁡−1(ex/a)+c\cos h^{- 1}(e^{x}/a) + \sec^{- 1}(e^{x}/a) + c

B

(1a)[cos⁡h−1(ex/a)+sec⁡−1(ex/a)]+c\left( \frac{1}{a} \right)\left\lbrack \cos h^{- 1}(e^{x}/a) + \sec^{- 1}(e^{x}/a) \right\rbrack + c

C

(1a)[cos⁡h−1(ex/a)+sec⁡−1(ex/a)]+c\left( \frac{1}{a} \right)\left\lbrack \cos h^{- 1}(e^{x}/a) + \sec^{- 1}(e^{x}/a) \right\rbrack + c

D

None of these

Answer

cos⁡h−1(ex/a)+sec⁡−1(ex/a)+c\cos h^{- 1}(e^{x}/a) + \sec^{- 1}(e^{x}/a) + c

Explanation

Solution

∫ex+aex−adx=\int \sqrt { \frac { e ^ { x } + a } { e ^ { x } - a } } d x = ∫ex+ae2x−a2dx\int_{}^{}{\frac{e^{x} + a}{\sqrt{e^{2x} - a^{2}}}dx} =∫exe2x−a2dx+∫ae2x−a2dx= \int \frac { e ^ { x } } { \sqrt { e ^ { 2 x } - a ^ { 2 } } } d x + \int \frac { a } { \sqrt { e ^ { 2 x } - a ^ { 2 } } } d x =∫ex/a(ex/a)2−12dx+∫1(ex/a)2−12dx= \int \frac { e ^ { x } / a } { \sqrt { \left( e ^ { x } / a \right) ^ { 2 } - 1 ^ { 2 } } } d x + \int \frac { 1 } { \sqrt { \left( e ^ { x } / a \right) ^ { 2 } - 1 ^ { 2 } } } d x

Put exa=t⇒exadx=dt\frac { e ^ { x } } { a } = t \Rightarrow \frac { e ^ { x } } { a } d x = d t =∫dtt2−1+∫dttt2−1=cos⁡h−1(ex/a)+sec⁡−1(ex/a)+c.= \int_{}^{}\frac{dt}{\sqrt{t^{2} - 1}} + \int_{}^{}\frac{dt}{t\sqrt{t^{2} - 1}} = \cos h^{- 1}(e^{x}/a) + \sec^{- 1}(e^{x}/a) + c.