Integral 2x^2 ln(x^2)

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• Apr 28th 2009, 02:38 PM
dianjoey
Integral 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 PM
skeeter
Quote:

Originally Posted by dianjoey
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)

$2x\ln(x^2) = 2x \cdot 2\ln{x} = 4x \ln{x}$

$4\int_1^e x\ln{x} \, dx$

integration by parts ...

$u = \ln{x}$

$dv = x \, dx$

proceed ...
• Apr 28th 2009, 03:35 PM
dianjoey
Hi Skeeter,

But the answer is $e^2+1$...
• Apr 28th 2009, 03:41 PM
skeeter
Quote:

Originally Posted by dianjoey
Hi Skeeter,

But the answer is $e^2+1$...

I agree ... you need to finish it.
• Apr 28th 2009, 03:55 PM
dianjoey
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 PM
dianjoey
I will show...

I provide the answer I want those who is willing to explain can check.
• Apr 28th 2009, 05:15 PM
redsoxfan325
Quote:

Originally Posted by skeeter
$2x\ln(x^2) = 2x \cdot 2\ln{x} = 4x \ln{x}$

$4\int_1^e x\ln{x} \, dx$

integration by parts ...

$u = \ln{x}$

$dv = x \, dx$

proceed ...

Now we have $4\cdot\frac{x^2}{2}\ln(x)-4\int\frac{x^2}{2}\cdot\frac{1}{x}\,dx = 2x^2\ln(x)-4\int\frac{x}{2}\,dx$

Can you finish?

Spoiler:
$=2x^2\ln(x)-x^2 = \left[x^2(2\ln(x)-1)\right]_1^e$ $= e^2(2\ln(e)-1) - 1^2(2\ln(1)-1) = e^2(2-1) - 1(0-1) = e^2+1$, as desired.
• Apr 28th 2009, 07:06 PM
dianjoey
Thx Redsoxfan325, very clear step and you way of solving the integral is easier to understand.
• Apr 28th 2009, 07:34 PM
mr fantastic
Quote:

Originally Posted by dianjoey
[snip]
Since ln (x^2) = 2 ln (x),

[snip]

As small point ..... ln(x^2) = 2 ln |x|.

But since you're working with x > 0 there is no difference between ln x and ln |x| in this question.