# How to find $\int{\frac{x}{\sqrt{x^2+1}}dx}$?

I started simplifying $$\int{\dfrac{x}{\sqrt{x^2+1}}dx}$$ but I always get this: $$\int{x(x^2+1)^{-1/2}dx}.$$

But I don't know how to follow by that way.

-
No matter how frustrating a mathematical problem can be, shit is inappropriate language here. – user2468 Apr 25 '12 at 20:35
@J.D. Sorry. I didn't know about it. – Garmen1778 Apr 25 '12 at 20:36
Have you tried substituting $u = x^2 + 1$? – user15464 Apr 25 '12 at 20:36
@ArturoMagidin Actually its not $$\int{x(x^2+1)^{-1}}$$. Its $$\int{x(x^2+1)^{-1/2}}$$. But thanks anyways. – Garmen1778 Apr 25 '12 at 20:40

Hint. Use the substitution $u=x^2+1$. Then $du = 2x\,dx$, so $x\,dx = \frac{1}{2}\,du$. Therefore, $$\frac{x}{\sqrt{x^2+1}}\,dx = \frac{\frac{1}{2}\,du}{\sqrt{u}} = \frac{1}{2}u^{-1/2}\,du.$$

-

Here's a hint: $$\int \frac{1}{\sqrt{1+x^2}}{\huge(}x\,dx{\huge)}.$$ If you don't know what that hints at, you should learn it. When you do understand what it hints at, you'll understand how to evaluate most integrals that are done by substitution.

-
Wouldn't it be better to write it as $$\int {\frac{1}{{\sqrt {1 + {x^2}} }}d\left( {\frac{{{x^2}}}{2}} \right)}$$ to make it a hint?? – Peter Tamaroff Apr 25 '12 at 20:58
@PeterTamaroff : I don't think so. The point is to tell the student what to look for when the student sees the integral. – Michael Hardy Apr 26 '12 at 12:51
@PeterTamaroff ......and you're giving the free of charge something that they should be learning how to do by themselves. The point of the hint was to help them learn how to do that step. Your version does that step for them. – Michael Hardy Apr 26 '12 at 15:01
I see. My idea was to point out to a substitution. I guess it'd have been more appropriate to put $$\int {\frac{1}{{\sqrt {1 + {x^2}} }}d\left( {\frac{{1 + {x^2}}}{2}} \right)}$$ – Peter Tamaroff Apr 26 '12 at 17:00

This looks like a problem for . . . u-substitution!

Let $u$ = $x^{2} + 1$. Thus, $du = 2x dx$.

$$\frac{1}{2} \int \frac{1}{\sqrt{u}} du$$

$$\frac{1}{2} \cdot {2\sqrt{u}}$$

Substituting $u$ back in for terms of $x$ and multiplying the $\frac {1}{2}$ by $2$, you are left with

$$\sqrt{x^{2} + 1}$$

-

Substituting $x^2 + 1 = y$, hence $xdx = \frac{1}{2}dy$. Therefore, the integral becomes $\int \frac{1}{2\sqrt{y}}dy$ which should be simple to evaluate.

-