# 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?

Take out $$e^x$$ from numerator and denominator as follows $$\int\dfrac{e^{2x}-1}{\sqrt{e^{3x}+e^x} } \ dx=\int\dfrac{e^x(e^{x}-e^{-x})}{\sqrt{e^{2x}(e^{x}+e^{-x})} } dx$$ $$=\int\dfrac{e^x(e^{x}-e^{-x})}{e^x\sqrt{e^{x}+e^{-x}} } dx$$ $$=\int\dfrac{(e^{x}-e^{-x})dx}{\sqrt{e^{x}+e^{-x}} }$$ $$=\int\dfrac{d(e^{x}+e^{-x})}{\sqrt{e^{x}+e^{-x}} }$$ $$=2\sqrt{e^{x}+e^{-x}}+C$$

• Interesingly enough, WolframAlpha cannot handle this without hypergeometric functions...+1 Jun 29, 2020 at 6:51
• @DonAntonio I'm getting roots and exponentials, though as usual it's not combining the root because it's worried $x$ might be complex. Jul 1, 2020 at 3:20

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 $$\theta$$ $$I=\frac 2{\sqrt{\sin(\theta)\cos(\theta)}}=\frac {2\sqrt 2}{\sqrt{\sin(2\theta)}}+C$$

$$\int\!\dfrac{e^{2x}-1}{\sqrt{e^{3x}+e^x}}\mathop{dx} =\int\!\dfrac{e^{x}-e^{-x}}{\sqrt{e^{x}+e^{-x}}}\mathop{dx} =\int\!\dfrac{2\sinh x}{\sqrt{2\cosh x}}\mathop{dx} =2\sqrt{2\cosh x} + C = 2\sqrt{e^{x}+e^{-x}} + C$$

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!

$$\int\dfrac{e^{2x}-1}{\sqrt{e^{3x}+e^x} } dx$$

$$=\int\dfrac{e^x(e^{x}-\frac{1}{e^x})}{\sqrt{e^{2x}(e^{x}+\frac1 {e^x})} } dx$$

$$=\int\dfrac{(e^{x}-e^{-x})}{\sqrt{e^{x}+e^{-x}} } dx$$

substitute $$e^x+e^{-x}=u$$, $$(e^x-e^{-x})dx=du$$,

$$=\int\frac{du}{\sqrt{u}}$$

$$=\frac{u^{-\frac12+1}}{-\frac12+1}$$

$$=2\sqrt{u}+c$$