$\int_{0}^{1} t^2 \sqrt{(1+4t^2)}dt$ solve

To compute the integral in question first consider that $$\cosh^2(x)-\sinh^2(x)=1,$$ so $$\cosh^2(x)=1+\sinh^2(x).$$

Making the substitution $t=\frac{\sinh(x)}{2},$ then $4t^2=\sinh^2(x),$ and we get $dt=\frac{\cosh(x)}{2}dx$

$$\int t^2 \sqrt{(1+4t^2)}dt=\frac{1}{8}\int{\sinh^2(x)}\sqrt{1+\sinh^2(x)}\cosh(x)dx$$

$$=\frac{1}{8}\int{\sinh^2(x)}\cosh^2(x)dx=\frac{1}{8}\left(\frac{1}{32}\sinh(4x)-4x\right) +C.$$

I leave the change of parameter for $t=0$ and $t=1$ for you to finish.