Question

\[\Huge \int \sqrt{4x^2+1} ~ dx\]

Answer 1

I don't know where to start...

Answer 2

trig sub

Answer 3

With what?

Answer 4

\[x={1\over2}\tan\theta\]

Answer 5

or \[
x={1\over2}\sinh t\\
dx={1\over2}\cosh t dt\\
I=\int \sqrt{\sinh^2t+1}{1\over2}\cosh tdt\\
I={1\over2}\int \cosh^2t dt\\
I={1\over2}\int {\cosh2t+1\over2}dt\\
\vdots
\]