Question

Please answer this question ......... ​

Answer 1

$I= \: _{0}∫^{ ln(13) } (\frac{ {e}^{x} \sqrt{ {e}^{x} - 1 } }{ {e}^{ x} + 3 } ) \: dx$

Let,

$\sqrt{e {}^{x} - 1 } = u$

$u {}^{2} = e {}^{x} - 1$

$2u \: du = {e}^{x} \: dx$

.

Now,

$I = 2 \:. \: _0∫^{ ln(13) } ( \frac{ {u}^{2} }{ {u}^{2} + 4} )du$

$I = 2 \: . \: _0∫^{ ln(13) } ( \frac{( {u}^{2} + 4) - 4}{( {u }^{2} + 4)} )du$

$= 2 \: . (_{2} ∫ ^{ ln(13) } du \: - 4 \: . \: \: _{0}∫^{2 \sqrt{3} } \frac{du}{ {u}^{2} + 4 } )$

=$2( \sqrt{ {e}^{x} - 1} )|_{0} ^{ ln(13) } - \frac{8}{2} \: \tan { }^{ - 1} ( \frac{ \sqrt{ {e}^{x} - 1} }{2} ) |_{0} ^{ ln(13) }$

=$2( \sqrt{ {e}^{ ln(13) } - 1 } - \sqrt{ {e}^{0} - 1} ) - 4 \: \tan^{ - 1} ( \frac{ \sqrt{ {e}^{ ln(13) } - 1 } }{2} ) - 4 \tan {}^{ - 1} (0)$

$I= 2 \sqrt{12} - 4 \tan ^{ - 1} ( \sqrt{3} )$

$I= 2 \sqrt{12} - 4 \frac{\pi}{3}$