Integral of x/sqrt(1-x^2). Question regarding this integral.

[jamesrb]

I was shown the answer and now I have a question in retrospect.
\(\displaystyle \int \frac{x}{\sqrt{1-x^{2}}}\)

So we need to do a substitution:
\(\displaystyle u=u=1-x^{2}\)
\(\displaystyle du=-2dx\)

Next:
\(\displaystyle -\frac{1}{2}\int \frac{1}{\sqrt{u}}\)

Next, the integral of \(\displaystyle \frac{1}{\sqrt{u}}=2\sqrt{u}\)

Why isn't it the integral of \(\displaystyle \frac{x}{\sqrt{u}}\)?

Where does the x in my numerator go? Something happened to it after the u substitution that I don't understand. The integral of x isn't 1, I know that, and it also isn't just x.

 

[MarkFL]

With the substitution:

\(\displaystyle u=1-x^2\)

you should have:

\(\displaystyle du=-2x\,dx\)

 

[jamesrb]

With the substitution:

\(\displaystyle u=1-x^2\)

you should have:

\(\displaystyle du=-2x\,dx\)

Oh yeah, I copied the answer incorrectly. I still can't see why it's not x/sqrtu though.

 

[lookagain]

I was shown the answer and now I have a question in retrospect.
\(\displaystyle \int \frac{x}{\sqrt{1-x^{2}}}\) . . . You're missing "dx" here.

So we need to do a substitution:
\(\displaystyle u=u=1-x^{2}\)
\(\displaystyle du=-2dx\)

Then dx = (-1/2)du.

Next:
\(\displaystyle -\frac{1}{2}\int \frac{1}{\sqrt{u}}\)

You're missing "du" here.

Next, the integral of \(\displaystyle \frac{1}{\sqrt{u}}=2\sqrt{u}\)

Why isn't it the integral of \(\displaystyle \frac{x}{\sqrt{u}}\)?

Where does the x in my numerator go? Something happened to it after the u substitution that I don't understand.
The integral of x isn't 1, I know that, and it also isn't just x.

\(\displaystyle \int\dfrac{x}{\sqrt{1 - x^2}}dx \)

(x)dx = \(\displaystyle \frac{-1}{2}du\)