1. ## Integration #1

$f(x) = xtan^{-1}x$

Find $f'(x)$ and hence find $\int_{0}^{1} tan^{-1}(x) dx$.

I can find $f'(x)$, but I don't know how to hence find the integral. We've covered (in the context of this forum) basic integration techniques in class, but not inverse tan. I looked up how to do it in general, but my confusion arises from the wording -- hence -- which suggests that I should be able to figure it out from the derivative. Any help is greatly appreciated.

2. ## Re: Integration #1

Originally Posted by Fratricide
$f(x) = xtan^{-1}x$

Find $f'(x)$ and hence find $\int_{0}^{1} tan^{-1}(x) dx$.

I can find $f'(x)$, but I don't know how to hence find the integral. We've covered (in the context of this forum) basic integration techniques in class, but not inverse tan. I looked up how to do it in general, but my confusion arises from the wording -- hence -- which suggests that I should be able to figure it out from the derivative. Any help is greatly appreciated.
do this by parts

if you know it that is

$\int udv= uv-\int vdu$

$u=tan^{-1}(x); du = \frac{1}{x^2+1}dx, dv=dx, v=x$

that's how I'd solve the integral

3. ## Re: Integration #1

Let's see:
f(x) = x*arctan(x) so
f'(x) = arctan(x) + x/(1+x^2)
Then,
arctan(x) = f'(x) - x/(1+x^2)
Integrating both sides with respect to x, we should have
S arctan(x)dx = f(x) - (1/2)ln(1+x^2)
Now, plug in x=0 and 1 and do the subtraction. xD