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
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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 integration by parts ... proceed ...
Hi Skeeter, But the answer is ...
Originally Posted by dianjoey Hi Skeeter, But the answer is ... I agree ... you need to finish it.
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)
I will show... I provide the answer I want those who is willing to explain can check.
Originally Posted by skeeter integration by parts ... proceed ... Now we have Can you finish? Spoiler: , as desired.
Thx Redsoxfan325, very clear step and you way of solving the integral is easier to understand.
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.
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