How can I integrate $\int\frac{e^{2x}-1}{\sqrt{e^{3x}+e^x} } \mathop{dx}$? 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.
 A: $$
\int\!\dfrac{e^{2x}-1}{\sqrt{e^{3x}+e^x}}\mathop{dx}
=\int\!\dfrac{e^{x}-e^{-x}}{\sqrt{e^{x}+e^{-x}}}\mathop{dx}
=\int\!\dfrac{2\sinh x}{\sqrt{2\cosh x}}\mathop{dx}
=2\sqrt{2\cosh x} + C = 2\sqrt{e^{x}+e^{-x}} + C
$$
A: Take out $e^x$ from numerator and denominator as follows
$$\int\dfrac{e^{2x}-1}{\sqrt{e^{3x}+e^x} } \ dx=\int\dfrac{e^x(e^{x}-e^{-x})}{\sqrt{e^{2x}(e^{x}+e^{-x})} } dx$$
$$=\int\dfrac{e^x(e^{x}-e^{-x})}{e^x\sqrt{e^{x}+e^{-x}} } dx$$
$$=\int\dfrac{(e^{x}-e^{-x})dx}{\sqrt{e^{x}+e^{-x}} } $$
$$=\int\dfrac{d(e^{x}+e^{-x})}{\sqrt{e^{x}+e^{-x}} } $$
$$=2\sqrt{e^{x}+e^{-x}}+C $$
A: I used the same steps you did follow but I stopped at
$$I=\int  \left(1-\cot ^2(\theta )\right) \sec (\theta )\sqrt{\tan (\theta )}\, d\theta$$
Rewrite it as
$$I={\displaystyle\int}\dfrac{\cos^2\left(\theta\right)-\sin^2\left(\theta\right)}{\cos^\frac{3}{2}\left(\theta\right)\sin^\frac{3}{2}\left(\theta\right)}\,d\theta$$ Now
$$u=\cos\left(\theta\right)\sin\left(\theta\right)\implies du=\cos^2\left(\theta\right)-\sin^2\left(\theta\right)\implies d\theta=\dfrac{du}{\cos^2\left(\theta\right)-\sin^2\left(\theta\right)}$$
$$I=\int\dfrac{du}{u^\frac{3}{2}}u==-\dfrac{2}{\sqrt{u}}+C$$ Back to $\theta$
$$I=\frac 2{\sqrt{\sin(\theta)\cos(\theta)}}=\frac {2\sqrt 2}{\sqrt{\sin(2\theta)}}+C$$
A: You are on right track. You can continue from here
$$=\int(\cot^{-3/2}\theta-\sqrt{\cot\theta} )\csc\theta d\theta $$
Substitute $\cot\theta=\frac{\cos\theta}{\sin\theta}$ & $\csc\theta=\frac1{\sin\theta}$
$$=\int\left(\frac{\sin\theta}{\cos\theta}\sqrt{\frac{\sin\theta}{\cos\theta}}-\sqrt{\frac{\cos\theta}{\sin\theta}} \right)\frac1{\sin\theta} d\theta $$
$$=\int\left(\frac{\sin^2\theta-\cos^2\theta}{\cos\theta\sqrt{\sin\theta\cos\theta}} \right)\frac1{\sin\theta} d\theta $$
$$=\int\frac{\left( \frac{1}{\cos^2\theta}-\frac{1}{\sin^2\theta}\right)}{\sqrt{\frac{1}{\sin\theta\cos\theta}}}d\theta$$
$$=\int\frac{( \sec^2\theta-cosec^2\theta)}{\sqrt{\tan\theta+\cot\theta}}d\theta $$
Let $\tan\theta+\cot\theta=t\implies (\sec^2\theta-cosec^2\theta)\ d\theta=dt$
$$=\int \frac{dt}{\sqrt {t}}$$
$$=2\sqrt {t}+C$$
Substitute $t=\tan\theta+\cot\theta$
$$=2\sqrt {\tan\theta+\cot\theta}+C$$
Substitute $\tan\theta=e^x$
$$=2\sqrt {e^x+e^{-x}}+C$$
Reached the answer. Cheers!
A: $$\int\dfrac{e^{2x}-1}{\sqrt{e^{3x}+e^x} } dx$$
$$=\int\dfrac{e^x(e^{x}-\frac{1}{e^x})}{\sqrt{e^{2x}(e^{x}+\frac1 {e^x})} } dx$$
$$=\int\dfrac{(e^{x}-e^{-x})}{\sqrt{e^{x}+e^{-x}} } dx$$
substitute $e^x+e^{-x}=u$, $(e^x-e^{-x})dx=du$,
$$=\int\frac{du}{\sqrt{u}}$$
$$=\frac{u^{-\frac12+1}}{-\frac12+1}$$
$$=2\sqrt{u}+c$$
A: $\displaystyle\int\frac{e^{2x}-1}{\sqrt{e^{3x}+e^{x}}}dx=\int\frac{e^{-x}(e^{2x}-1)}{\sqrt{e^{-2x}(e^{3x}+e^{x})}}dx=\displaystyle\int\frac{e^{x}-e^{-x}}{\sqrt{e^{x}+e^{-x}}}dx$
$\displaystyle z=e^{x}+e^{-x}\Rightarrow dz=(e^{x}-e^{-x})dx$
$\displaystyle\int\frac{e^{2x}-1}{\sqrt{e^{3x}+e^{x}}}dx=\int\frac{dz}{\sqrt{z}}=2\sqrt{z}+c$
$\displaystyle\int\frac{e^{2x}-1}{\sqrt{e^{3x}+e^{x}}}dx=2\sqrt{e^{x}+e^{-x}}+c$
