Integral of (lnx)^2/x, calculus u-substitution

[@scitwi9164]

00:36 Instead of solving for `dx`, you could just observe that you have `dx/x` on the right side, which matches the `dx/x` in your integral, so you could replace this part with `du` inside the integral right away. After replacing also `(ln x)²` with `u²`, you get the integral of `u²·du` without the need of solving for `dx`. This also applies for many other of your examples. Just a little shortcut ;)

[@shikhaparjapati2859]

Wow!sir thnku so much love from India👌👌👌

[@blackpenredpen]

Integral of (lnx)^2/x

[@joaomiguelbarros7950]

YOU ARE THE BEST. THANK YOU.

[@youtubeviewerFF]

thank you kind human ♥

[@garystewart7322]

Thank you very much buddy!

[@Hogojub]

Yay i did it! I learned this kind of math only because of your videos

[@Hogojub]

I did it with once with u-sub and once with IBP :)

[@anthonyjh02]

1:48 where does the final du go? It just disappears?

[@carultch]

When you carry out an integral, the differential (the "d whatever" term) disappears by definition. Given that f(x) is the derivative of F(x), this means when you integrate: integral f(x) dx The result is: F(x) + C In this case, we're integrating u^2 relative to u. This means f(x) = u^2, and F(x) is a function such that F'(x) = u^2. The power rule allows us to find F(x), by increasing the exponent by 1, and having the new exponent join the coefficient as the denominator, giving us 1/3*u^3. We of course add the +C to allow this to represent all possible functions whose derivative in the u-world is u^2.

[@tarek_talab]

thanks a lot

[@anarosef]

Thank you so much !

[@hannahs_music]

Thank you

[@sivabharathisivam5411]

Thank you very much.😊

[@ahmedoumakhir8536]

thank you sir

[@edimoperondijr5098]

Thank's for the video!!

[@magomorlin]

thanks

[@dahakabarkley3389]

Thx

[@777loveforever]

Muchas gracias!!

[@procerpat9223]

you can do this in your head

[@i_am_anxious02]

x=e^u dx=e^u•du Integral of u^2•e^(-u)•e^udu Integral of u^2du u^3/3 ln(x)^3/3

[@lesliesilva5774]

jiji justo lo que buscaba

[@stevenjaimes9578]

Leslie Silva pero esa integral esta mal bb

[@mohammedhmad5261]

Find the integral of( (lnx+1)^2)/x. I'm not know sulution

[@mohammedhmad5261]

Please help me

[@ernestschoenmakers8181]

@@mohammedhmad5261 Coming to the rescue, you have the following integral: I = int of (ln(x)+1)^2/x*dx, let's open up the brackets then you'll get: int of (ln^2(x)+2*ln(x)+1)/x*dx, now split the integral into 3 terms: int of (ln^2(x))/x*dx+ int of 2*ln(x)/x*dx+int of (1/x)*dx, know:(1/x)*dx=d(ln(x)) so we apply this to the integrals: int of ln^2(x)d(ln(x))+int of 2*ln(x)d(ln(x))+int of d(ln(x))= I = (1/3)*ln^3(x) + ln^2(x) + lnIxI + C, this is the solution.

[@farazriyaz9078]

You can integrate it as shown above or you can take u=lnx+1, which I think would make things easier. Then, dx=xdu You get, ∫((lnx+1)²/x)dx = ∫u²du Thus, ∫u²du = u³/3 + C = (lnx+1)³/3 + C Which you can expand as such: (lnx)³/3 + (lnx)² + lnx + C