qazplmty
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When integrating lnx2 would you do it by parts or substitution? I think i know how to do it by parts but apparently you can do it by substitution too?
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TenOfThem
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(Original post by ZarqaaX)
When integrating lnx2 would you do it by parts or substitution? I think i know how to do it by parts but apparently you can do it by substitution too?
Parts
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Semaj18
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The integral of lnx is xlnx - x
So as lnx^2 is 2lnx, the integral of lnx^2 is 2 times the integral of lnx.
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james22
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Is that ln(x^2) or (ln(x))^2?
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qazplmty
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(Original post by james22)
Is that ln(x^2) or (ln(x))^2?
(ln(x))^2, apparently its done by substitution and we should learn it
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user2020user
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(Original post by ZarqaaX)
(ln(x))^2, apparently its done by substitution and we should learn it
I = \displaystyle \int \left[ \ln x \right]^2 \ dxNow that you mention it, you could rewrite the integral as:

I = \displaystyle \int x \ \dfrac{ \left[ \ln x \right]^2 }{x}\ dx

and proceed by using the substitution u=\ln x.

You might need to do a bit of rearranging of the substitution to make x the subject then use integration by parts (twice perhaps) on an exponential term multiplied by a polynomial.

(Original post by james22)
Is that ln(x^2) or (ln(x))^2?
I'm pretty sure it's the latter, otherwise we could simply use the power rule for logarithms and show that the integral of the former is simply 2 lots of that of the integral of the natural logarithm.
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qazplmty
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(Original post by Khallil)
I = \displaystyle \int \left[ \ln x \right]^2 \ dxNow that you mention it, you could rewrite the integral as:

I = \displaystyle \int x \ \dfrac{ \left[ \ln x \right]^2 }{x}\ dx

and proceed by using the substitution u=\ln x.

You might need to do a bit of rearranging of the substitution to make x the subject then use integration by parts (twice perhaps) on an exponential term multiplied by a polynomial.
Ohhh thank you! Ive gotten to the part where i substituted it as intergration of u^2 (e^u du) but what do i do with the e^u do i intergrate that too? By substitution or parts? Or leave it?
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user2020user
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(Original post by ZarqaaX)
Ohhh thank you! Ive gotten to the part where i substituted it as intergration of u^2 (e^u du) but what do i do with the e^u do i intergrate that too? By substitution or parts? Or leave it?
By parts (twice)
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qazplmty
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(Original post by Khallil)
By parts (twice)
Wouldnt i just get e^u again? Because e^u integrated is e^u right?
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james22
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(Original post by ZarqaaX)
Wouldnt i just get e^u again? Because e^u integrated is e^u right?
Yes, but you are differentiating the u^2 so it becomes u. Do it again and the u becomes 1 and you can just integrate e^u then.
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user2020user
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Substitutions
U = u^2 \implies \dfrac{dU}{du} = 2u

\dfrac{dV}{du} = e^u \implies V = e^u

(Original post by ZarqaaX)
Wouldnt i just get e^u again? Because e^u integrated is e^u right?
\displaystyle \begin{aligned} I & = \int e^u u^2\ du \\ & \overset{I.B.P}= u^2 e^u - \underbrace{ \int 2u e^u\ du }_{\alpha} \end{aligned}

That's what you get when you use IBP the first time. Try it again and see if you can reduce the u in \alpha down to a constant.
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qazplmty
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Oh now i understand, thank you so much everyone for helping me I fully understand it now, thanks for explaining it to me
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