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∫ √ E X − 1 D X - Mathematics

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Question

\[\int\sqrt {e^x- 1}  \text{dx}\] 
Sum
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Solution

\[\int\sqrt{e^x - 1}dx\]
\[\text{Let e}^x - 1 = t^2 \]
\[ \Rightarrow e^x = t^2 + 1\]
\[ e^x = \text{2t }\frac{dt}{dx}\]
`dx = {2t   dt}/{e^x} `
`dx = {2t   dt}/{t^2 + 1} `
\[Now, \int\sqrt{e^x - 1}dx\]
`  = ∫   {  t  . 2t   dt}/{t^2 + 1} `
`  =2  ∫   {  t^2   dt}/{t^2 + 1} `
\[ = 2\ ∫ \left( \frac{t^2 + 1 - 1}{t^2 + 1} \right)dt \]
\[ = 2\ ∫ dt - 2\int\frac{dt}{t^2 + 1}\]
\[ = 2t - 2 \tan^{- 1} \left( t \right) + C\]
\[ = 2\sqrt{e^x - 1} - 2 \tan^{- 1} \left( \sqrt{e^x - 1} \right) + C\]

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Chapter 19: Indefinite Integrals - Exercise 19.09 [Page 59]

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RD Sharma Mathematics [English] Class 12
Chapter 19 Indefinite Integrals
Exercise 19.09 | Q 66 | Page 59

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