- #1

Pythagorean

Gold Member

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The problem: [tex] \int \,x \arctan \,x \,dx [/tex]

(i'm assuming tan to the negative first power is the same as arctan)

the answer: [tex] \frac{1+x^2}{2}arctan x - \frac{x}{2} [/tex]

My process so far:

I first looked at dividing by parts with u=x and dv= arctan x dx. That got really messy trying to integrate tan, so I flipped them over.

What erks me is I don't feel like I'm violating any rules, I'm just getting a different answer.

So I set u=arctan x and dv= x dx

so [itex] du=\frac{1}{x^2+1} [/itex] and [itex] v= \frac{1}{2}x^2 [/itex]

which leaves me with:

(using uv - [itex] \int [/itex]dvu)

[tex] \frac{1}{2}x^2 arctan x - \frac{1}{2}\int \frac{x^2}{x^2+1} [/tex]

I use long division on that last term so that I have:

[tex]\int 1 - \frac{1}{x^2+1} [/tex]

which results in

[tex] \frac{1}{2}x^2 arctan x - \frac{1}{2}arctan x - \frac{1}{2}x [/tex]