Indefinite Integral

try x= a tan y

\hspace{-16}\mathbf{\int\sqrt{x+\sqrt{x^2+a^2}}\;dx}$\\\\\\ Let $\mathbf{\sqrt{x^2+a^2}+x=e^{2t}\Leftrightarrow \frac{x+\sqrt{x^2+a^2}}{\sqrt{x^2+a^2}}dx=2e^{2t}dt}\\\\\\ \mathbf{\frac{e^{2t}}{\sqrt{x^2+a^2}}dx=2e^{2t}dt\Leftrightarrow dx=2.\sqrt{x^2+a^2}dt}$\\\\\\ Now $\mathbf{\sqrt{x^2+a^2}+x=e^{2t}..............................(1)}$\\\\\\ Multiply both side by $\mathbf{\sqrt{x^2+a^2}-x},\;$ We Get \\\\\\ $\mathbf{\left(\sqrt{x^2+a^2}+x\right).\left(\sqrt{x^2+a^2}-x\right)=e^{2t}.\left(\sqrt{x^2+a^2}-x\right)}$\\\\\\ $\mathbf{\sqrt{x^2+a^2}-x=a^2.e^{-2t}.............................(2)}$\\\\\\ So Add equation...$\mathbf{(1)}$ and $\mathbf{(2)},\;$ we Get\\\\\\ $\mathbf{2\sqrt{x^2+a^2}=\left(e^{2t}+a^2.e^{-2t}\right)}$\\\\\\ Now $\mathbf{dx=2\sqrt{x^2+a^2}dt=\left(e^{2t}+a^2.e^{-2t}\right)dt}$\\\\\\ So $\mathbf{\int\sqrt{x+\sqrt{x^2+a^2}}\;dx=\int e^t.\left(e^{2t}+a^2.e^{-2t}\right)dt}$\\\\\\

\hspace{-16}\mathbf{=\int e^{3t}dt+a^2.\int e^{-t}dt}$\\\\\\ $\mathbf{=\frac{1}{3}.e^{3t}-a^2.e^{-t}+C}$\\\\\\ $\mathbf{=\frac{1}{3}.\left(\sqrt{x^2+a^2}+x\right)^\frac{3}{2}-a^2.\left(\sqrt{x^2+a^2}+x\right)^\frac{-1}{2}+C}$

What hinted at the substitution by e^2t ?

may be becz it is always positive and d(e^x) = e^x dx

I thought of x=atanα, followed by cosα=1√2(cosθ).