Find the value of integration following : $\int x^2\sqrt{x^2+1}dx$ Find the value of integration following :
$$\int x^2\sqrt{x^2+1} dx$$
Let $x=\sinh(t)$ and $dx=\cosh(t) \,dt$. So, $$I=\int x^2\sqrt{x^2+1}\ dx=\int \sinh^2(t)\cosh^2(t)\,dt=\frac 14 \int \sinh^2(2t)\,dt=\frac 18 \int (\cosh(4t)-1)\,dt$$ $$I=\frac 1{32}\sinh(4t)-\frac t8+C$$
$$
\frac{1}{4}(x^2+1)^{\frac{3}{2}}-\frac{x}{8}\sqrt{x^2+1}-\frac{1}{8}\mathrm{arcsinh}(x)
$$
Differentiating the above will give you $x^2\sqrt{x^2+1}$, proving it is indeed the antiderivative.