# Indefinite integral of $\int \frac{x^3}{\sqrt{x^2+1}} \,\text{d}x$.

Can you please provide any sort of hint or suggestion on how to find the following indefinite integral?

$$\int\frac{x^3}{\sqrt{x^2+1}}\text{d}x$$

I tried substituting everything but it didn't work. I also tried trigonometric substitution but I couldn't find any valid trigonometric identity for $\sqrt{\cos^2(x)+1}$ or $\sqrt{\sin^2(x)+1}$.

We have

$$\int \frac{x^3}{\sqrt{x^2 + 1}}\, dx = \int x \frac{(x^2 + 1) - 1}{\sqrt{x^2 + 1}}\, dx = \int x\sqrt{x^2 + 1}\, dx - \int \frac{x}{\sqrt{x^2 + 1}}\, dx.$$

Now use the $u$-substitution $u = x^2 + 1$ to get

$$\frac{1}{2}\int \sqrt{u}\, du - \frac{1}{2}\int \frac{du}{\sqrt{u}} =...$$

• it would a lot simpler if you do $u^2 = x^2 + 1.$ – abel Mar 15 '15 at 20:50
• Yes abel, but the OP may not be comfortable with the less obvious substitution. – kobe Mar 15 '15 at 20:52
• kobe, you seem to know a wide variety of mathematical areas. what did you do to get there? – abel Mar 15 '15 at 20:54
• @abel it's mainly from studying various mathematical disciplines. – kobe Mar 15 '15 at 22:56
• thanks. can you elaborate on the disciplines? how long have you been doing? – abel Mar 15 '15 at 22:59

Hint: Make the substitution $x^2+1=v^2$

The answer will come out as $\left(\dfrac{v^3}{3}-v+C\right)$ where C is the integration constant.

• You should leave this as a comment. – GohP.iHan Apr 21 '15 at 4:42
• @GohP.iHan, that doesn't give me reputation, so.... :3 – Prasun Biswas Apr 21 '15 at 5:19

$$\int\frac{x^2}{\sqrt{x^2+1}}\Big(x\,dx\Big) = \frac 1 2 \int \frac{u}{\sqrt{u+1}} \, du = \frac 1 2 \int \frac{w^2-1}{w} \Big(2w\, dw\Big) = \cdots$$ ($w=\sqrt{u+1}$, so $w^2-1=u$ and then differentiating both sides of the last equality we get $2w\,dw=du$.)