Pythagorean
Sep21-04, 11:12 PM
so, I know the integration by formula for this, since it's in my handy book of formulas, but i'm required to do this by parts, and then use a technique from the lesson to integrate it further.
The problem: \int \,x \arctan \,x \,dx
(i'm assuming tan to the negative first power is the same as arctan)
the answer: \frac{1+x^2}{2}arctan x - \frac{x}{2}
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 du=\frac{1}{x^2+1} and v= \frac{1}{2}x^2
which leaves me with:
(using uv - \int dvu)
\frac{1}{2}x^2 arctan x - \frac{1}{2}\int \frac{x^2}{x^2+1}
I use long division on that last term so that I have:
\int 1 - \frac{1}{x^2+1}
which results in
\frac{1}{2}x^2 arctan x - \frac{1}{2}arctan x - \frac{1}{2}x
The problem: \int \,x \arctan \,x \,dx
(i'm assuming tan to the negative first power is the same as arctan)
the answer: \frac{1+x^2}{2}arctan x - \frac{x}{2}
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 du=\frac{1}{x^2+1} and v= \frac{1}{2}x^2
which leaves me with:
(using uv - \int dvu)
\frac{1}{2}x^2 arctan x - \frac{1}{2}\int \frac{x^2}{x^2+1}
I use long division on that last term so that I have:
\int 1 - \frac{1}{x^2+1}
which results in
\frac{1}{2}x^2 arctan x - \frac{1}{2}arctan x - \frac{1}{2}x