How can I solve this $\int\dfrac{1}{\sqrt{5e^{-2x}+4e^{-x}+1} } \mathop{dx}=?$

Question

How can I solve this $$\int\dfrac{1}{\sqrt{5e^{-2x}+4e^{-x}+1} } \mathop{dx}=?$$

My attempt:

I substituted $e^{-x}=t$, $-e^{-x}\ dx=dt$, $dx=-\dfrac{dt}{t}$ $$\int\dfrac{1}{\sqrt{5t^2+4t+1 } }\left(-\dfrac{dt}{t}\right) $$

$$=-\dfrac{1}{\sqrt{5}}\int\dfrac{1}{t\sqrt{t^2+\dfrac45t+\dfrac15 } }dt $$ $$=-\dfrac{1}{\sqrt{5}}\int\dfrac{1}{t\sqrt{\left(t+\dfrac25\right)^2+\dfrac{1}{25} } }dt $$

I substituted $t+\dfrac25=u \ $,$\ dt=du$ $$=-\dfrac{1}{\sqrt{5}}\int\dfrac{1}{\left(u-\dfrac25\right)\sqrt{u^2+\dfrac{1}{25} } }du $$

It seems to me that I am not in the right direction.

Please help me solve this integral. Thanks.

Answer 1

Multiply $e^x$ to numerator and denominator & proceed as follows $$\int\dfrac{1}{\sqrt{5e^{-2x}+4e^{-x}+1}} dx $$ $$=\int\frac{e^x}{\sqrt{5+4e^{x}+e^{2x}} } dx $$ $$=\int\frac{e^x\ dx}{\sqrt{1+(e^{x}+2)^2} }$$ $$=\int\frac{d(e^x+2)}{\sqrt{(e^{x}+2)^2+1}}$$ $$=\sinh^{-1}(e^{x}+2) +C$$

Answer 2

Continue with \begin{align} &\int\dfrac{1}{\sqrt{5t^2+4t+1 } }(-\dfrac{dt}{t} ) \overset{u=\frac1t} =\int\dfrac{du}{\sqrt{(u+2)^2 +1} } =\sinh^{-1}(u+2) +C\\ \end{align}

Answer 3

$$\begin{align*} \int \frac{dx}{\sqrt{5e^{-2x}+4e^{-x}+1}} &= \int \frac{e^x\,dx}{\sqrt{5+4e^x+e^{2x}}} \\ &= \int \frac{dy}{\sqrt{5+4y+y^2}} & y=e^x\\ &= \int \frac{\frac12 \frac{5-4z+z^2}{(z-2)^2}}{z+\frac12 \frac{5-z^2}{z-2}} \,dz & y=\frac12 \frac{5-z^2}{z-2}\\ &= \int \frac{dz}{z-2} \end{align*}$$