Partial fraction decomposition for a trig function

Question

I have to admit I couldn't find a way to deal with this problem. I am a newbie in mathematics.

$$\int \frac{1}{e^{2x}\cosh(x)^3}dx$$

The only idea I have is to substitute and use the formula below.

$$\int \frac{1}{\cosh^2}dx = \tanh(x)+c $$

Answer 1

Using $$\cosh\left(x\right)=\frac{e^{x}+e^{-x}}{2} $$ we have $$I=\int\frac{1}{e^{2x}\cosh^{3}\left(x\right)}dx=8\int\frac{e^{x}}{\left(e^{2x}+1\right)^{3}}dx $$ now put $e^{x}=u $ to get $$=8\int\frac{1}{\left(u^{2}+1\right)^{3}}du $$ and $u=\tan\left(v\right),\, du=\sec^{2}\left(v\right)dv $ $$=8\int\frac{1}{\left(\tan^{2}\left(v\right)+1\right)^{3}\cos^{2}\left(v\right)}dv=8\int\cos^{4}\left(v\right)dv $$ which we can rewrite, using power reduction formula, as $$=\int\left(3+4\cos\left(2v\right)+\cos\left(4v\right)\right)dx=3v+2\sin\left(r\right)+\frac{\sin\left(s\right)}{4}+C $$ where $r=2v $ and $s=4v $. Now you have to substitute back.

Answer 2

By setting $x=\log t$ we have: $$\int\frac{dx}{e^{2x}\cosh^3 x}=\int\frac{8 e^{x}\,dx}{e^{6x}+3e^{4x}+3e^{2x}+1}=\int\frac{dt}{(t^2+1)^3}=\frac{1}{8}\left(\frac{5t+3t^3}{(1+t^2)^2}+3\arctan t\right).$$

The last equality follows from: $$\frac{1}{t^2+1}=\frac{1}{2i}\left(\frac{1}{t-i}-\frac{1}{t+i}\right)$$ that leads to: $$\frac{1}{(t^2+1)^2} = -\frac{1}{4}\left(\frac{1}{(t-i)^2}+\frac{1}{(t+i)^2}-\frac{2}{t^2+1}\right)$$

$$ \frac{1}{(t^2+1)^3} = \frac{i}{8}\left(\frac{1}{(t-i)^3}-\frac{1}{(t+i)^3}+\frac{6i}{(t^2+1)^2}\right).$$