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प्रश्न
\[\int\sqrt {e^x- 1} \text{dx}\]
योग
उत्तर
\[\int\sqrt{e^x - 1}dx\]
\[\text{Let e}^x - 1 = t^2 \]
\[ \Rightarrow e^x = t^2 + 1\]
\[ e^x = \text{2t }\frac{dt}{dx}\]
`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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