I love how I think I am getting the hang of calc 2 and run into silly problems like this. It's one of those classes that seems deceptively easy at first and then slams you on your head.
Working on a Calc II assignment and found the concepts used to set up the problem extremely helpful for a more complex version. Thanks for doing what you do!
I solved the problem by choosing the inverse log aka e^x and let x = e^t and lnx = t So dx = e^t dt Now we are integrating t²e^t dt Results in t²e^t - 2te^t + 2e^t + C using the Tabular Method. We know, t=lnx So the integral of lnx^2 dx = xln²x - 2xlnx + 2x + C And results in the same answer! 🙂
@@TheOrganicChemistryTutor Can I ask when to use U-substitution or Integration by parts, like how do you identify that particular problem is solved by that technique. Thanks
Because you need the derivative of the inner function, to already be existing in the original integrand. This is trivial when the inner function is just a linear function of x, because you can ALWAYS make that happen by multiplying by 1 in a fancy way, producing a new constant in the integrand, and putting its reciprocal out in front. However, it isn't as trivial when the inner function isn't linear. This is the same reason why it is easy u-substitution to integrate x*e^(-x^2) dx, but not easy to integrate e^(-x^2) dx without the factor of x out in front. We do use the reverse chain rule to turn ln(x)^2 into u^2, but we also need to produce the derivative elsewhere in the integrand. We also need to differentiate u relative to x, so that we can produce du to replace dx. du = 1/x dx Solve for dx: dx = x du Replace dx in the original integral, to rewrite it as: integral u^2 * x du In order to proceed, we need this to be completely in the u-world. It's NOT OK for x to be straggling like this, since x itself is a function of u, and not a constant. We solve our definition of u for x, to replace x with e^u. integral u^2 * e^u du Now we have an integral that is completely in the u-world, and it is legal for us to proceed from here. This is a simple ender for integration by parts, where u^2 is differentiated to zero.
Late response, i know, but the reason we don't integrate by substitution is because when you take u=lnx you will be left with a du/x in the integration because the derivative for lnx is 1/x. Whilst when you integrate by parts you don't get that.
thank you for your work can i please ask why did you put 2 in front of the second integral in 2:53 ? i mean the first integral has the 2 already but why you did put it again ? thanx again :D
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I love how I think I am getting the hang of calc 2 and run into silly problems like this. It's one of those classes that seems deceptively easy at first and then slams you on your head.
I agree, fellow Jimmy
I am getting definitely getting slammed!
Right??
Exactly i just failed my test lol
Well said
Working on a Calc II assignment and found the concepts used to set up the problem extremely helpful for a more complex version. Thanks for doing what you do!
thank you sir. there are a lot of non obvious integrals that help you solve a breadth of problems that youve shown me.
I solved the problem by choosing the inverse log aka e^x and let x = e^t and lnx = t
So dx = e^t dt
Now we are integrating t²e^t dt
Results in t²e^t - 2te^t + 2e^t + C using the Tabular Method.
We know, t=lnx
So the integral of lnx^2 dx = xln²x - 2xlnx + 2x + C
And results in the same answer! 🙂
Thank you for showing these specific examples 👌
You are the best I learn almost everything from you🧘🏻♂️👑
U can put lnx=t, then dx=xdt and x can be written as e^t
Overall the integration is t².(e^t)dt
Dude thank you so much. I have been trying to solve this problem, but I can't Soo thank you Soo much again
Thank you for your work
You're welcome
@@TheOrganicChemistryTutor Can I ask when to use U-substitution or Integration by parts, like how do you identify that particular problem is solved by that technique. Thanks
@@paranoidparadigm886 got left on read for three years
what a boss man I love this guy. what a cool guy
why cant we just integrate (lnx)^2 by reverse chain rule?
Because you need the derivative of the inner function, to already be existing in the original integrand. This is trivial when the inner function is just a linear function of x, because you can ALWAYS make that happen by multiplying by 1 in a fancy way, producing a new constant in the integrand, and putting its reciprocal out in front. However, it isn't as trivial when the inner function isn't linear. This is the same reason why it is easy u-substitution to integrate x*e^(-x^2) dx, but not easy to integrate e^(-x^2) dx without the factor of x out in front.
We do use the reverse chain rule to turn ln(x)^2 into u^2, but we also need to produce the derivative elsewhere in the integrand. We also need to differentiate u relative to x, so that we can produce du to replace dx.
du = 1/x dx
Solve for dx:
dx = x du
Replace dx in the original integral, to rewrite it as:
integral u^2 * x du
In order to proceed, we need this to be completely in the u-world. It's NOT OK for x to be straggling like this, since x itself is a function of u, and not a constant. We solve our definition of u for x, to replace x with e^u.
integral u^2 * e^u du
Now we have an integral that is completely in the u-world, and it is legal for us to proceed from here. This is a simple ender for integration by parts, where u^2 is differentiated to zero.
Hiii, can you explain why integration by parts is used here? My first thought was to integrate by substitution with u= lnx as sub
Late response, i know, but the reason we don't integrate by substitution is because when you take u=lnx you will be left with a du/x in the integration because the derivative for lnx is 1/x. Whilst when you integrate by parts you don't get that.
Thanks but the similarity of u and v in your handwriting makes this unnecessarily confusing
Godamn I never thought of it that way. Thanks so much
Mükemmelsiniz hocam ❤
Thank you this helped so much
Why the fuck is the sign of integration suddenly a factor in the integral? How does this work at all? I am so confused.
my first idea looking at the problem is u substitution why can't we use it can anyone explain it
The Best!
thank you mate
Thanks!!!
No Problem
thank you for your work
can i please ask why did you put 2 in front of the second integral in 2:53 ?
i mean the first integral has the 2 already but why you did put it again ?
thanx again :D
He's distributing the 2 to get rid of the brackets.
hey thank you a lot
when I use derivative on the result, it gives me 2dx. why though?
Thank youu
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0:56 you took dedicate of u here but after that you took integration of v whys that
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