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 $\displaystyle 2x\ln(x^2) = 2x \cdot 2\ln{x} = 4x \ln{x}$ $\displaystyle 4\int_1^e x\ln{x} \, dx$ integration by parts ... $\displaystyle u = \ln{x} $ $\displaystyle dv = x \, dx $ proceed ...
Hi Skeeter, But the answer is $\displaystyle e^2+1$...
Originally Posted by dianjoey Hi Skeeter, But the answer is $\displaystyle e^2+1$... 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 $\displaystyle 2x\ln(x^2) = 2x \cdot 2\ln{x} = 4x \ln{x}$ $\displaystyle 4\int_1^e x\ln{x} \, dx$ integration by parts ... $\displaystyle u = \ln{x} $ $\displaystyle dv = x \, dx $ proceed ... Now we have $\displaystyle 4\cdot\frac{x^2}{2}\ln(x)-4\int\frac{x^2}{2}\cdot\frac{1}{x}\,dx = 2x^2\ln(x)-4\int\frac{x}{2}\,dx$ Can you finish? Spoiler: $\displaystyle =2x^2\ln(x)-x^2 = \left[x^2(2\ln(x)-1)\right]_1^e$ $\displaystyle = e^2(2\ln(e)-1) - 1^2(2\ln(1)-1) = e^2(2-1) - 1(0-1) = e^2+1$, 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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