Integral of sqrt((x^3-1)/x^11) (substitution + substitution)
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- เผยแพร่เมื่อ 3 ต.ค. 2024
- Integral of sqrt((x^3-1)/x^11) - How to integrate it step by step using the substitution method!
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thankyou so much i was getting bald trying to solve this exact problem 😭
💪😀
great solution sir thank u keep it up
Thank you! 😊😊
can you please do the integral of sqrt(x/x^3+1). tks.
Hi! Here you have the solution:
Integral of sqrt(x/(x^3+1)) dx =
= Integral of x^(1/2)/sqrt(x^3+1) dx =
= Integral of 1/sqrt(x^3+1) x^(1/2) dx =
Substitution:
u = x^(3/2) ==> u^2 = x^3
du = (3/2)x^(1/2) dx = 1/2u dx ==> (2/3) du = x^(1/2) dx
= Integral of 1/sqrt(1+u^2) (2/3) du =
= (2/3)Integral of 1/sqrt(1+u^2) du =
= (2/3) th-cam.com/video/Dp2jkPXU9hw/w-d-xo.html =
= (2/3)ln|sqrt(1+u^2) + u| =
= (2/3)ln|sqrt(1+x^3) + x^(3/2)| + C
Hope it helped! ;-D
I am failing to comprehend were sqrt 4 came from.
Hi! Do you mean sqrt(u)? Sorry for my bad handwritting...
@@IntegralsForYou I meant in minute 1:05, sqrt x to the power of 4.
@@cm6995sqrt of x^4 is the same as x^2 .Becasue x^4 can be written as the square of( x^2). so when we take root of square of (x^2) it will cancel out and we will get x^2
( i hope this is not too late )
sir, I am taking x^3 common and x^11 common respectively, then taking outside 1/x^4, then taking 1-1/x^3 = t, my answer is same, but it is 1/3 in place of 2/9
Oh...let's see:
Integral of sqrt((x^3-1)/x^11) dx =
= Integral of sqrt( (x^3-1)/(x^3)*(x^8) ) dx =
= Integral of sqrt( (1/x^8)(x^3-1)/(x^3) ) dx =
= Integral of (1/x^4)*sqrt( (x^3-1)/(x^3) ) dx =
= (1/x^4)*Integral of (1/x^4)*sqrt( x^3/x^3 - 1/x^3 ) dx =
= Integral of sqrt( 1 - 1/x^3 ) (1/x^4)dx =
Substitution:
u = 1 - 1/x^3 = 1 - x^(-3)
du = (0 - (-3)x^(-4))dx = (3/x^4)dx ==> (1/3)du = (1/x^4)dx
= Integral of sqrt(u) (1/3)du =
= (1/3)*Integral of sqrt(u) du =
= (1/3)*Integral of u^(1/2) du =
= (1/3)*u^(3/2)/(3/2) =
= (1/3)*(2/3)*u^(3/2) =
= (2/9)*u^(3/2) =
= (2/9)*(1-1/x^3)^(3/2) + C
I think you forgot the (3/2) dividing the u^(3/2) in the last steps... Hope it helped! ;-D
@@IntegralsForYou yes sir, I did a mistake in taking the power down. Thnx a lot
@@halofreak3644 My pleasure! 😉
what about the integral of ((2x+1)/(x+2)) ^ (1/3) plz
Hi! This is a very long integral... I cannot write the answer here or make a video now, but I suggest you to start doing a u=((2x+1)/(x+2))^(1/3) substitution...
@@IntegralsForYou Yes, I did that before, but I did not reach a conclusion. If there is time, try it plz
nice
Thanks! ;-D
6:10 can you please explain why did you cancel minus ?
Hi! The integral of f'(x)[f(x)]^n is equal to [f(x)]^(n+1)/(n+1).
We need the derivative of f(x) in front of [f(x)]^n. In our case, f(t)=1-t and its derivative is f'(t)=-1
(-1/3)Integral of (1-t)^(1/2) dt =
= (1/3)Integral of (-1)(1-t)^(1/2) dt =
= (1/3)(1-t)^(3/2)/(3/2)
Can you please integrate √x^3
Integral of √x^3 dx =
= Integral of x^(3/2) dx =
= x^(3/2 + 1)/(3/2 + 1) =
= x^(3/2 + 2/2)/(3/2 + 2/2) =
= x^(5/2)/(5/2) =
= (2/5)x^(5/2) + C
😉
@@IntegralsForYou thank you sir ji
@Hariom Chaubey You're welcome! 😊