$\int_{0}^{1} t^2 \sqrt{(1+4t^2)}dt$ solve $$\int_{0}^{1} t^2 \sqrt{(1+4t^2)}dt$$
my attempt
$$t = \frac{1}{2}\tan(u)$$
$$dt = \frac{1}{2}\sec^2(u)du\\$$
$$\begin{align}
\int_{0}^{1} t^2 \sqrt{(1+4t^2)}dt&=\int_{0}^{1} \frac{\tan^2(u)}{4} \sqrt{1+\tan^2(u)}\frac{1}{2}\sec^2(u)du\\
&=\frac{1}{8}\int_{0}^{1} \tan^2(u)\sec^{3}(u)du\\
&= \frac{1}{8} \int_{0}^{1} (\sec^2(u) - 1)(\sec^{3}(u))du\\
&=\frac{1}{8}\int_{0}^{1} \sec^5(u)du - \frac{1}{8} \int_{0}^{1}\sec^3(u)du
\end{align}$$
what now?

Hint.  I would say
$\int \sec^5u\,du = \int\frac{\cos u\,du}{\cos^6u}=\int\frac{\cos u\,du}{(1-\sin^2u)^3}$ and after substitution
$\sin u = v,\quad \cos u \,du = dv$
it will be an integral part of the rational function.
Edit:
I will add: The given integral is the so-called binomial integral with a specific solution. See
http://www.nabla.hr/CL-IndefIntegralB5.htm
