I am trying to solve the following irrational integral
$I=\int \sqrt{1+4t^2}\,dt $
Doing the change: $ t=\frac{1}{2}\sinh(x)$ , $dt= \frac{1}{2}\cosh(x)$
$I=\int \sqrt{1+\sinh^2x}(\frac{1}{2}\cosh x)\,dx =\int \sqrt{\cosh^2x}(\frac{1} {2}\cosh x)\,dx = \frac{1}{2}\int{\cosh^2x}\,dx$
Doing $\cosh x=\frac{e^x+e^{-x}}{2}$
$I=\frac{1}{2}\int{ (\frac{e^x+e^{-x}}{2}})^2 \,dx$$=\frac{1}{8}\int{ (e^{2x}+e^{-2x}+2}) \,dx$ $=\frac{1}{16} (e^{2x}-e^{-2x}+4x)$
from $ t=\frac{1}{2}\sinh(x)$ -> $x= arcsinh(2t)$ and $4t=e^x-e^{-x}$ so $16t^2 +2=e^{2x}-e^{-2x}$
then $I=\frac{1}{16} (e^{2x}-e^{-2x}+4x)=\frac{1}{16} (16t^2 +2+4arcsinh(2t))$$
I saw this solved somewhere else using another method, so I know the $16t^2 +2 $ part of the answer is wrong, However I can't spot my mistake. Can you spot it?