integration by parts

[roadrunner]

I am having trouble completing this problem:

Integrate xarctanx dx: I let u = arctanx, du = 1/(1 + x squared)

dv = xdx, v = xsquared/2

Using the parts rule I get: (xsquared/2)arctanx - 1/2 integral xsquared/(1+xsq) dx

I can't seem to complet the part of this problem after the subtraction sign. I tried letting u = 1/(1+x sq) and dv = x sq but that didn't work I don't think.

Then I let u = x, du - dx; v = 1/2(2x/1+x sa) dx but I thwarted there also.

Any suggestions? I don't think I'm supposed to use trig substitution as we haven't gotten to that yet. What do you think?[/code]

 

[daon]

I am having trouble completing this problem:

Integrate xarctanx dx: I let u = arctanx, du = 1/(1 + x squared)

dv = xdx, v = xsquared/2

Using the parts rule I get: (xsquared/2)arctanx - 1/2 integral xsquared/(1+xsq) dx

I can't seem to complet the part of this problem after the subtraction sign. I tried letting u = 1/(1+x sq) and dv = x sq but that didn't work I don't think.

Then I let u = x, du - dx; v = 1/2(2x/1+x sa) dx but I thwarted there also.

Any suggestions? I don't think I'm supposed to use trig substitution as we haven't gotten to that yet. What do you think?[/code]

\(\displaystyle \L\\ \int{x arctanxdx\)

You let u=arctanx and dv = xdx

du = \(\displaystyle \frac{dx}{1+x^2}\), and v=\(\displaystyle \frac{x^2}{2}\)

So \(\displaystyle \L\\ uv - \int vdu\) = \(\displaystyle \L\\ \frac{x^2arctanx}{2} - \frac{1}{2}\int \frac{x^2dx}{1+x^2}\)

\(\displaystyle \L\\ = \frac{x^2arctanx}{2} - \frac{1}{2}\int \frac{(x^2 + 1 - 1)dx}{1+x^2}\)

\(\displaystyle \L\\ = \frac{x^2arctanx}{2} - \frac{1}{2}\int 1dx - \frac{1}{2}\int\frac{(-1)dx}{1+x^2}\)

Where you had a problem, you needed to split up the fraction. You could have divided the x^2 by (x^2+1) or added and subtracted 1 like I did. Can you finish from there?

 

[roadrunner]

Yes, I can finish the problem now. I got stuck thinking I had to continue using parts. Thanks a lot for your help.