find f'(x), when f(x) is the integral of tan(t)/t

[green_tea]

Hi!
I have to find f '(x) when

\(\displaystyle f(x)=\int_{x^{2}}^{arctan(x)}\frac{tan(t)}{t}dt\), x>0

I thought that I could find the undefined integral of \(\displaystyle \frac{tan(t)}{t}\) first, then "put in" arctan(x) and x^2 and then derive the whole thing, but I cant integrate it..
Is it even possible? I tried many methods but I can't find it. Then I tried to solve it online on this site
http://integrals.wolfram.com/index.jsp?expr=x^3/(x^2+1)&random=false
and it said that most likely no formula exist. :?

So is there any other way to find the f '(x) ?

 

[PAULK]

Try the rule for differentiating an integral:
When you have:

{u(x)
| g(t) dt = G(u) - G(v),
}v(x)

where, of course, G' = g.

Now, how would you differentiate G(u(x)) - G(v(x))? You already know:

1. How to differentiate G.
2. The chain rule.

That should do it. You don't have to know how to integrate tan t/t.

 

[galactus]

Remember:

\(\displaystyle \frac{d}{dx}\int_{f(x)}^{g(x)}f(t)dt=f(g(x))g'(x)-f(h(x))h'(x)\)

In this case, \(\displaystyle g(x)=tan^{-1}(x), \;\ h(x)=x^{2}\)