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

Question

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.

Answer 1

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

Answer 2

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.

Answer 3

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

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

So now, your integral is:

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

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

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

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

Answer 4

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.