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).