Integral of x*ln(x+1) (by parts)

[@IntegralsForYou]

🔍 𝐀𝐫𝐞 𝐲𝐨𝐮 𝐥𝐨𝐨𝐤𝐢𝐧𝐠 𝐟𝐨𝐫 𝐚 𝐩𝐚𝐫𝐭𝐢𝐜𝐮𝐥𝐚𝐫 𝐢𝐧𝐭𝐞𝐠𝐫𝐚𝐥? 𝐅𝐢𝐧𝐝 𝐢𝐭 𝐰𝐢𝐭𝐡 𝐭𝐡𝐞 𝐢𝐧𝐭𝐞𝐠𝐫𝐚𝐥 𝐬𝐞𝐚𝐫𝐜𝐡𝐞𝐫: ► Integral searcher 👉integralsforyou.com/integral-searcher 🎓 𝐇𝐚𝐯𝐞 𝐲𝐨𝐮 𝐣𝐮𝐬𝐭 𝐥𝐞𝐚𝐫𝐧𝐞𝐝 𝐚𝐧 𝐢𝐧𝐭𝐞𝐠𝐫𝐚𝐭𝐢𝐨𝐧 𝐦𝐞𝐭𝐡𝐨𝐝? 𝐅𝐢𝐧𝐝 𝐞𝐚𝐬𝐲, 𝐦𝐞𝐝𝐢𝐮𝐦 𝐚𝐧𝐝 𝐡𝐢𝐠𝐡 𝐥𝐞𝐯𝐞𝐥 𝐞𝐱𝐚𝐦𝐩𝐥𝐞𝐬 𝐡𝐞𝐫𝐞: ► Integration by parts 👉integralsforyou.com/integration-methods/integration-by-parts ► Integration by substitution 👉integralsforyou.com/integration-methods/integration-by-substitution ► Integration by trig substitution 👉integralsforyou.com/integration-methods/integration-by-trig-substitution ► Integration by Weierstrass substitution 👉integralsforyou.com/integration-methods/integration-by-weierstrass-substitution ► Integration by partial fraction decomposition 👉integralsforyou.com/integration-methods/integration-by-partial-fraction-decomposition 👋 𝐅𝐨𝐥𝐥𝐨𝐰 @𝐢𝐧𝐭𝐞𝐠𝐫𝐚𝐥𝐬𝐟𝐨𝐫𝐲𝐨𝐮 𝐟𝐨𝐫 𝐚 𝐝𝐚𝐢𝐥𝐲 𝐢𝐧𝐭𝐞𝐠𝐫𝐚𝐥! 😉 📸 instagram.com/integralsforyou/

[@maicariuionel6667]

Thanx dude,passed my exam with this,love you!

[@IntegralsForYou]

Thank you for telling me it, you don't imagine how much I smile when I see comments like yours! Congratulations! 💪💪

[@maicariuionel6667]

@@IntegralsForYou thank you again,and i hope you have a great day! ❤✌

[@IntegralsForYou]

@Maicariu Ionel Thank you! ❤️

[@ishank2577]

Bro can you add voice explaination

[@red-sv2qf]

2:25 where did u get the integral of x from :/?

[@IntegralsForYou]

Hi! It comes from simplifying (x+1)(x-1)/(x+1) to (x-1). Here you have the details: Integral of x^2/(x+1) dx = = Integral of (x+1)(x-1)/(x+1) dx + Integral of 1/(x+1) dx = = Integral of (x+1)(x-1)/(x+1) dx + Integral of 1/(x+1) dx = = Integral of (x-1) dx + Integral of 1/(x+1) dx = = Integral of x dx - Integral of 1 dx + Integral of 1/(x+1) dx = = x^2/2 - x + ln|x+1| Hope it helped! 💪

[@red-sv2qf]

thank you so much @@IntegralsForYou

[@IntegralsForYou]

@@red-sv2qf My pleasure! ❤

[@俞豆陽]

Thank you , you are the best

[@IntegralsForYou]

Thank you! 😊

[@yashvardhansharma4945]

Great video!!!! thanks for this!!!!!

[@IntegralsForYou]

My pleasure! 😎

[@pradyushkumarjena7265]

Thanks for solving it . Why you don't speak in your lecture?

[@IntegralsForYou]

Hi, I know that it would be easier for you if I talked but I prefer to let you think why am I doing each step and if you don't understand it you can ask me in a comment. In my opinion, it is the best way to practice integration 😊

[@pradyushkumarjena7265]

OK sir

[@amandeepbaghiana4510]

@@IntegralsForYou It would also help if you just said why you were doing the things you're doing, on my end if I am confused I can simply pause the video and contemplate the problem.

[@miguelblx18.]

Gracias pa te amo

[@IntegralsForYou]

Y yo te amo más si me ayudas a hacer crecer mi nuevo canal sobre derivadas jeje... www.youtube.com/@DerivativesForYou Gracias! ❤❤

[@godofgamers3884]

I’m sorry but how are we supposed to truly understand if you’re just writing without explaining anything?

[@IntegralsForYou]

Hi, I know that it would be easier for you if I talked but I prefer to let you think why am I doing each step and if you don't understand it you can ask me in a comment. In my opinion, it is the best way to practice integration 😉

[@reaj3404]

can you explain what you did at 1:16. I didn't quite understand however other than that nice job thank you.

[@IntegralsForYou]

Hi! I am doing a polynom division. I think this video will help you: kzbin.info/www/bejne/o37akJqVedh5es0

[@juliocastellanos2257]

Nice, gracias!

[@IntegralsForYou]

De nada! Un saludo! ;-D

[@mahatnicolas8612]

Thank men

[@IntegralsForYou]

You're welcome! 😊👍

[@isaacnavarro4614]

thank you

[@IntegralsForYou]

You're welcome! ;-D

[@aziminamdar5271]

By substituting X=e^u-1 Problem becomes less complicated with exponential component with an algebraic component . We could simply use UV rule Am I right ?? Plz let me know

[@IntegralsForYou]

Hi! Yes, you can use that trick: Integral of x*ln(x+1) dx = Substitution: x = e^t - 1 ==> x+1=e^t ==> ln(x+1) = t dx = (e^t)dt = Integral of (e^t - 1)*ln(e^t - 1 + 1) (e^t)dt = = Integral of (e^t - 1)*ln(e^t) (e^t)dt = = Integral of (e^t - 1)*t (e^t)dt = = Integral of t*(e^2t - e^t) dt = Parts: Integral of u dv = uv - Integral of v du u = t ==> du = dt dv = (e^2t - e^t)dt ==> v = (1/2)e^(2t) - e^t = t*((1/2)e^(2t)-e^t) - Integral of ( (1/2)e^2t - e^t ) dt = = t*(e^t)( (1/2)e^(t)-1 ) - ( (1/2)(1/2)e^2t - e^t ) = = t*(e^t)( (1/2)e^(t)-1 ) - (e^t)((1/4)e^t - 1) = = ln(x+1)(x+1)*( (1/2)(x+1) - 1 ) - (x+1)*( (1/4)(x+1) - 1 ) = = ln(x+1)(x+1)*( (1/2)x + 1/2 - 1 ) - (x+1)*( (1/4)x + 1/4 - 1 ) = = ln(x+1)(x+1)*( (1/2)x - 1/2 ) - (x+1)*( (1/4)x - 3/4 ) = = (1/2)ln(x+1)(x+1)(x-1) - (1/4)(x+1)*(x-3) = = (1/2)(x^2-1)ln(x+1) - (1/4)(x^2-3x+x-3) = = (1/2)(x^2-1)ln(x+1) - (1/4)(x^2-2x-3) = = (1/2)(x^2-1)ln(x+1) - (1/4)(x^2-2x-3) = = (x^2/2)ln(x+1) - (1/2)ln(x+1) - (1/4)x^2 + (1/2)x + 3/4 + C = = (x^2/2)ln(x+1) - (1/2)ln(x+1) - (1/4)x^2 + (1/2)x + C' (where C'=3/4 + C is also a constant) ;-D

[@manhba3050]

Integral of x.ln^2(x+1)dx = ??? 🙏🙏🙏✏📚📖

[@IntegralsForYou]

Hi! Here you have the solution: Integral of x*ln^2(x+1) dx = Substitution: t = x+1 ==> t-1=x dt = dx = Integral of (t-1)ln^2(t) dt = Parts: Integral of u dv = uv - Integral of v du u = ln^2(t) ==> du = 2*ln(t)/t dt dv = (t-1)dt ==> v = t^2/2 - t = ln^2(t)*(t^2/2 - t) - Integral of (t^2/2 - t)2*ln(t)/t dt = = ln^2(t)*(t^2/2 - t) - Integral of (t - 2)ln(t) dt = Parts: Integral of u dv = uv - Integral of v du u = ln(t) ==> du = 1/t dt dv = (t-2)dt ==> v = t^2/2 - 2t = ln^2(t)*(t^2/2 - t) - [ ln(t)(t^2/2 - 2t) - Integral of (t^2/2 - 2t)(1/t) dt ] = = ln^2(t)*(t^2/2 - t) - ln(t)(t^2/2 - 2t) + Integral of (t/2 - 2) dt = = ln^2(t)*(t^2/2 - t) - ln(t)(t^2/2 - 2t) + (t^2/4 - 2t) = = (t/4)[ ln^2(t)*(2t - 4) - ln(t)(2t - 8) + (t - 8) ] = = ((x+1)/4)[ ln^2(x+1)*(2(x+1) - 4) - ln(x+1)(2(x+1) - 8) + (x+1-8) ] = = ((x+1)/4)[ (2x-2)ln^2(x+1) - (2x-6)ln(x+1) + x - 7 ] + C ;-D

[@ahmedalhisaie7698]

mate, you've lost me at 1:11 can't get anything after that. my question almost the same only I have x to the power 3

[@IntegralsForYou]

Hi Ahdmed Alhisaie, why do you have x^3? From 1:11 I am doing a polynomial division

[@IntegralsForYou]

For the integral of x^2/x+1 dx you can also do the next substitution: Integral of x^2/x+1 dx = Substitution: t=x+1 ==> t-1=x dt=dx = Integral of (t-1)^2/t dt = = Integral of (t^2 - 2t + 1)/t dt = = Integral of (t^2/t - 2t/t + 1/t) dt = = Integral of (t - 2 + 1/t) dt = = t^2/2 - 2t + ln|t| = = (x+1)^2/2 - 2(x+1) + ln|x+1|

[@ahmedalhisaie7698]

Integrals ForYou no it just the question that I have, is little bit different, however yes I didn't understand from this time 1:11 I'm right now trying to learn polynomial division it seems little bit complicated

[@IntegralsForYou]

Good luck with polynomial division, it seems complicated but from the third example you will see it is always the same :D I have some integrals in this playlist where I do polynomial division if you want: kzbin.info/aero/PLpfQkODxXi49YODSGByJH2O7dVRgzPKe2

[@ahmedalhisaie7698]

mate , integrate this one please . x^2 ln(1+x)dx

[@shibani187]

Hi sir .. what about xlog(1+2x)...After integrating by parts .. I stopped at x^2/1+2x... Plz help

[@IntegralsForYou]

Integral of x*ln(1+2x) dx = = Integral of ln(1+2x) x dx = Parts: Integral of u dv = uv - Integral of v du u = ln(1+2x) ==> du = 2/(1+2x) dx dv = x dx ==> v = x^2/2 = ln(1+2x)*(x^2/2) - Integral of (x^2/2)(2/(1+2x)) dx = = (x^2/2)ln(1+2x) - Integral of x^2/(1+2x) dx = Polynomial division: D = dq + r x^2 |__2x+1 -x^2 -(1/2)x ------- ----------------- (1/2)x - 1/4 / -(1/2)x +(1/2)x + 1/4 ----------------------- / 1/4 x^2 = (1+2x)( (1/2)x - 1/4 ) + 1/4 = (1/4)( (1+2x)(2x-1) + 1) ==> x^2/(1+2x) = = (1/4)( (1+2x)(2x-1) + 1)/(1+2x) = = (1/4)( (1+2x)(2x-1)/(1+2x) + 1/(1+2x) )= = (1/4)( (2x-1) + 1/(1+2x) )= = (1/4)( 2x - 1 + 1/(1+2x) ) ==> = (x^2/2)ln(1+2x) - Integral of x^2/(1+2x) dx = = (x^2/2)ln(1+2x) - Integral of (1/4)( 2x - 1 + 1/(1+2x) ) dx = = (x^2/2)ln(1+2x) - (1/4)Integral of ( 2x - 1 + 1/(1+2x) ) dx = = (x^2/2)ln(1+2x) - (1/4)[ Integral of 2x dx - Integral of dx + Integral of 1/(1+2x)) dx ] = = (x^2/2)ln(1+2x) - (1/4)[ 2*Integral of dx - Integral of dx + (1/2)*Integral of 2/(1+2x)) dx ] = = (x^2/2)ln(1+2x) - (1/4)[ 2*x^2/2 - 1 + (1/2)ln|(1+2x)| ] = = (x^2/2)ln(1+2x) - (1/4)[ x^2 - 1 - (1/2)ln|(1+2x)| ] = = (x^2/2)ln(1+2x) - x^2/4 + 1/4 - (1/8)ln|(1+2x)| = = (1/8)[ (4x^2)ln(1+2x) - 2x^2 + 2 - ln|(1+2x)| ] = = (1/8)[ (4x^2 - 1)ln(1+2x) - 2x^2 + 2 ] + C

[@deajapon]

dv should be (x dx) and not just x

[@IntegralsForYou]

Hi, Felicia! I agree, I forgot to write it...

[@summy2780]

Bro, you missed some thing

[@IntegralsForYou]

The sound? 🙃

[@summy2780]

@@IntegralsForYou If I was right, it is 3/4

[@IntegralsForYou]

@@summy2780 Hi! I need more details... could you tell me your complete solution or how did you get there? Thanks!

[@fatimazohra4893]

pleas i need this integrale ln(x)/(1+x^2) from 0 to +infinity

[@romario21]

Use bigger font !!!Use a sharp Pen !!!

[@IntegralsForYou]

Hi! Newer videos have better quality, take a look 😉

[@tohzhenhong5183]

change pen pls and ur 1 is like the same as ur other signs , totally cannot see what u write

[@nicolasramos4107]

Para mi esta mal

[@IntegralsForYou]

Hola, Nicolás, por qué lo dices? Igual simplemente tienes una solución tan válida como la mía... 😉

[@joshuajumper8898]

This answer is wrong. -ln(x+1) / 2 is incorrect it should be +3/4 + C at the end. The error was made when he factored his x squared and gave his 1 a whole new body (nun/denomin) instead of cancelling it.

[@IntegralsForYou]

Hi Joshua! I don't understand how did you get your +3/4. Can you please tell me how did you do it? However, I have reviewed and I think my answer is correct....

[@duffyhasswag6453]

Problem and neither of you can give me the right answer

[@simonahalep2885]

It doesn't matter Joshua The admin's answer is write that 3/4 you're talking about will go as the constant that you're saying c.

[@enhace15anos.83]

where the hell did you get 3/4?💀

[@rnasrinisha2495]

wrong answer

[@IntegralsForYou]

Hi R Nasrinisha, I don't agree with you. Maybe you have a different answer but equivalent to mine. However, you should say the right answer if you think mine is wrong...

[@AppleUploader]

Thanks for explaining it. Isn’t that the whole point of trying to teach? Thumbs down.

[@mitko17]

Well, it doesn't say tutorial soo...