Find the value of integration following : $\int x^2\sqrt{x^2+1}dx$ Find the value of integration following :
$$\int x^2\sqrt{x^2+1} dx$$
 A: Let $x=\sinh(t)$ and $dx=\cosh(t) \,dt$. So, $$I=\int x^2\sqrt{x^2+1}\ dx=\int \sinh^2(t)\cosh^2(t)\,dt=\frac 14 \int \sinh^2(2t)\,dt=\frac 18 \int (\cosh(4t)-1)\,dt$$ $$I=\frac 1{32}\sinh(4t)-\frac t8+C$$
A: Let $x=\tan u$ and $dx= \sec^2u du$ 
$\displaystyle\int x^2\sqrt{x^2+1} dx=\int \tan^2 u\sqrt{\tan^2 u+1} \sec^2 u du=\int (\tan^2 u \sec^3 u )du=\int (1-\sec^2 u)(\sec^3 u)du=\int (\sec^3 u-\sec^5 u )du$
Now, use the reduction formula for $\displaystyle\int \sec^m u du=\frac{\sec^{m-2}u \tan u}{m-1}+\frac{m-2}{m-1}\int \sec^{m-2}udu$.
A: You can alternately apply the following tranform:
Let,
$$x=\frac{1}{2}\Big(t-\frac{1}{t}\Big)$$
So,
$$dx=\frac{1}{2}\Big(1+\frac{1}{t^2}\Big)dt$$
and
$$\sqrt{x^2+1}=\Big(t+\frac{1}{t}\Big)$$
Thus, your integral becomes:
$$I=\int\frac{1}{4}\Big(t-\frac{1}{t}\Big)^2\frac{1}{2}\Big(1+\frac{1}{t^2}\Big)\Big(t+\frac{1}{t}\Big)dt$$ which can be easily-though perhaps somewhat tediously-evaluated.
A: $$
\frac{1}{4}(x^2+1)^{\frac{3}{2}}-\frac{x}{8}\sqrt{x^2+1}-\frac{1}{8}\mathrm{arcsinh}(x)
$$
Differentiating the above will give you $x^2\sqrt{x^2+1}$, proving it is indeed the antiderivative.
