I got problem to integral ln (x^2)...

As the primitive of In (x) = x(ln (x) -1)

I was wondering... Since ln (x^2) = 2 ln (x), could it work in this way to solve an integral???

The answer is: e^2 + 1

Thx in adv for your explanation (Rock)

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- Apr 28th 2009, 02:38 PMdianjoeyIntegral 2x^2 ln(x^2)
I got problem to integral ln (x^2)...

As the primitive of In (x) = x(ln (x) -1)

I was wondering... Since ln (x^2) = 2 ln (x), could it work in this way to solve an integral???

The answer is: e^2 + 1

Thx in adv for your explanation (Rock) - Apr 28th 2009, 03:21 PMskeeter
- Apr 28th 2009, 03:35 PMdianjoey
Hi Skeeter,

But the answer is $\displaystyle e^2+1$... - Apr 28th 2009, 03:41 PMskeeter
- Apr 28th 2009, 03:55 PMdianjoey
Skeeter,

I am just re-doing my old exercises, I have all steps to get the answer, but I just don't understand how to integral ln (x^2) - Apr 28th 2009, 04:06 PMdianjoey
I will show...

I provide the answer I want those who is willing to explain can check. - Apr 28th 2009, 05:15 PMredsoxfan325
- Apr 28th 2009, 07:06 PMdianjoey
Thx Redsoxfan325, very clear step and you way of solving the integral is easier to understand.

- Apr 28th 2009, 07:34 PMmr fantastic