Need help with log integral with change of variable.

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Opus89

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One of my homework problems is the integral of x/(x^2+1) dx. I'm terribly sorry that I don't know how to make the integral look like its supposed to but I hope you can get the idea. I'm having a mental block. I know i need to let u= x^2 + 1 and du= 2xdx. I just don't know what to do after that. I know I will end up with a 1/2 on the outside of the integrand but I'm having trouble getting that far. I just started calc 2 last week and I did well with the change of variable problems in calc 1(last sem) but I can't for the life of me figure this out. I just need someone to point me in the right direction. Thanks
 
Differeniate \(\displaystyle y=\ln (x^2 +1)\).
That should tell what to do next.
 
Opus89 said:
One of my homework problems is the integral of x/(x^2+1) dx. I'm terribly sorry that I don't know how to make the integral look like its supposed to but I hope you can get the idea. I'm having a mental block. I know i need to let u= x^2 + 1 and du= 2xdx. I just don't know what to do after that. I know I will end up with a 1/2 on the outside of the integrand but I'm having trouble getting that far. I just started calc 2 last week and I did well with the change of variable problems in calc 1(last sem) but I can't for the life of me figure this out. I just need someone to point me in the right direction. Thanks
Click to expand...

\(\displaystyle \int\frac{x\, dx}{x^2+1}\)

Next step is to substitute into the given integral

\(\displaystyle =\frac{1}{2}\cdot\int\frac{2x\, dx \, (=\, du)}{(x^2+1)\, (= \, u)}\)

Now do you see the next step....
 
The thing that seems to hang my up is I'm not sure why you need to multiply and divide by 2. I just don't know what purpose that serves. I know you have to, I just don't know why.
 
\(\displaystyle \int\frac{x}{x^{2}+1}dx\)

Make the sub \(\displaystyle u=x^{2}+1, \;\ du=2xdx, \;\ \frac{1}{2}du=xdx\)

See now where the 1/2 comes from?. Just make your subs. You got that far and that's good.

Replace the \(\displaystyle x^{2}+1\) with a u and the xdx with the \(\displaystyle \frac{1}{2}du\)

After that is done, all you have is a easy integral.
 
Yep. That's it.

\(\displaystyle \frac{1}{2}\int\frac{1}{u}du\)

Now, it's easy, is it not?.
 
Yea i guess so. 1/u du = ln |u| + c. Since, u = x^2 + 1, we have ln |x^2 + 1| + c. And we have the 1/2 which makes it (1/2) ln |x^2 + 1| +c. For some reason, i was unable to recall the fact that you could pull the fraction constant (1/2) out of just the numerator. I thought the 1 needed to be in the numerator and the 2 needed to be in the denominator if that makes any sense at all. Sort of like this: 1/ 2u.
 
Opus89 said:
… For some reason, i was unable to recall the fact that you could pull the fraction constant (1/2) out of just the numerator. I thought the 1 needed to be in the numerator and the 2 needed to be in the denominator …
Click to expand...


Are you talking about the following step?

\(\displaystyle du = 2x \; dx\)

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

If so, then the reason that you cannot recall might be that you're thinking of factoring, instead of simply dividing both sides by 2.

If not, then I don't know what you're trying to say.

 
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