Finding antiderivative of x / sqrt[x^2 + 1] by substitution

[roam]

Hello!

This is my question:

Find the antiderivative of $\displaystyle \frac{x}{\sqrt{x^2 +1}}$

Here is my working, please correct me if I am wrong:

Using the method of substitution;
$\displaystyle u^2 = x^2 +1$
$\displaystyle 2udu = 2xdx$

$\displaystyle x = \sqrt{(u^2 -1)}$

$\displaystyle \int \frac{\sqrt{(u^2 -1)}}{u}\times \frac{u}{\sqrt{(u^2 -1)}} du$

$\displaystyle \frac{1}{u} \times u du$

Substituting back we have:
$\displaystyle \frac{1}{\sqrt{x^2 +1}} \times \sqrt{x^2 +1} du$

$\displaystyle = 1 du$

So, is that it? am I right?

I'd appreciate it if you could please help me.

Regards,

 

[o_O]

Re: Antiderivatives!

You could've gotten from:
$\displaystyle \int \left(\frac{\sqrt{u^{2} - 1}}{u} \cdot \frac{u}{\sqrt{u^{2} - 1}}\right) du$

directly to:

$\displaystyle \int du$

by direct cancellation.

Anyway, just integrate the final expression and resubstitute in AFTER you integrate (that's the whole point of making the subs after all).