How can I evaluate this integral $$\int\dfrac{e^{2x}-1}{\sqrt{e^{3x}+e^x} } \mathop{dx}=\;\;?$$
My attempt:
I tried using substitution $e^x=\tan\theta$, $e^x\ dx=\sec^2\theta\ d\theta$, $dx=\sec\theta \csc\theta \ d\theta.$ $$\int\dfrac{\tan^2\theta-1}{\sqrt{\tan^3\theta+\tan\theta } } \sec\theta \csc\theta\ d\theta $$
$$=\int\dfrac{\tan^2\theta-1}{\sec\theta\sqrt{\tan\theta } } \sec\theta \csc\theta d\theta. $$ I used $\tan\theta= \dfrac{1}{\cot\theta}$ $$=\int\dfrac{1-\cot^2\theta}{\cot^{3/2}\theta }\csc\theta d\theta $$ $$=\int(\cot^{-3/2}\theta-\sqrt{\cot\theta} )\csc\theta d\theta. $$ I got stuck here. I can't see whether further substitution will work or not. Will integration by parts work?
Please help me solve this integral. I am learning calculus. Thank in advance.
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TShiong
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