Trigonometric / Mixed Integral Involving Substitution

[StopWatch]

Homework Statement

integrate xarctanx/(x^2 + 1)^2

Homework Equations

Integration by parts possibly? I was attempting to do it without integration by parts because we went over this in tutorial and my TA couldn't solve it properly, hence why I'm here.

The Attempt at a Solution

I thought I could sub u = xarctanx so that du = 1/(x^2 + 1) dx but then I'm still left with one of the two x^2 + 1's. Wolfram gave me usinucosu du as the integral after substitution was complete and I feel like knowing how this was possible will be useful if I see any integrals like this in the future (rather than having to try integrating by parts).

 

[SammyS]

Homework Statement

integrate xarctanx/(x^2 + 1)^2

Homework Equations

Integration by parts possibly? I was attempting to do it without integration by parts because we went over this in tutorial and my TA couldn't solve it properly, hence why I'm here.

The Attempt at a Solution

I thought I could sub u = xarctanx so that du = 1/(x^2 + 1) dx ...

To find the derivative of x arctan(x), you need to use the product rule, so your result is incorrect. In fact, your result is the derivative of arctan(x) by itself.

...but then I'm still left with one of the two x^2 + 1's. Wolfram gave me usinucosu du as the integral after substitution was complete and I feel like knowing how this was possible will be useful if I see any integrals like this in the future (rather than having to try integrating by parts).

Did you understand any of what WolframAlpha did?

 

[StopWatch]

I didn't understand much of what wolfram did, except in principle. I made a mistake though: It substituted u = arctanx alone, so du = 1/(x^2 + 1) which led it to the conclusion that this was equivalent to the integral usinucosu du. You're right that I will eventually have to use parts, and I think I can handle that if I just know how it got this substitution.

 

[StopWatch]

But I personally don't see anywhere that usinucosu would have come from.

 

[StopWatch]

Anyone?

 

[SammyS]

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

Let $\displaystyle u=\arctan(x)\quad\to\quad du=\frac{1}{x^2+1}dx\,.$

That takes care of everything except $\displaystyle \frac{x}{x^2+1}\,.$

Also, x = tan(u), so that gives us: $\displaystyle \int\frac{\tan(u)\,(u)}{(\tan^2(u)+1)}\,du\,.$

Use the trig identity tan²(u) + 1 = sec²(u) .

Then simplify.

 

[StopWatch]

Thank you so much, I really appreciate it! I need to make sure I realize that arctan(x) = u also says something about x (which should have been rather obvious) in the future.