1. ## Integration

I have a differentiable function, $\displaystyle f$

Find $\displaystyle \int xf '(x^2) \, dx$

2. Hello,
Originally Posted by WWTL@WHL

I have a differentiable function, $\displaystyle f$

Find $\displaystyle \int xf '(x^2) \, dx$
Substitute $\displaystyle t=x^2$

3. Thanks for the quick reply, Moo.

I'm probably being really silly, but I just can't finish it off.

Using $\displaystyle t = x^2$

$\displaystyle \int \frac{t^{-\frac{1}{2}}}{2} f '(t) \, dt$

Using substitution by parts...

$\displaystyle \int \frac{t^{-\frac{1}{2}}}{2} f '(t) \, dt = \frac{t^{-\frac{1}{2}}}{2} f'(t) + \frac{1}{4} \int t^{-\frac{3}{2}} f(t)$

Any idea?

Thanks.

EDIT: Hang on for 5 mins, I think I've spotted my mistake.

4. Ok, so my previous post was garbage.

I've got to:

$\displaystyle \frac{1}{2} \int f'(t) \, dt$

$\displaystyle = \frac{1}{2} f(t)$

$\displaystyle = \frac{1}{2} f(x^2)$

is this right?

5. Originally Posted by WWTL@WHL
Ok, so my previous post was garbage.

I've got to:

$\displaystyle \frac{1}{2} \int f'(t) \, dt$

$\displaystyle = \frac{1}{2} f(t)$

$\displaystyle = \frac{1}{2} f(x^2)$

is this right?

Except that you forgot the integration constant !

6. Constant, Schmanstant

Excellent. Thanks for your help, Moo.