Integral Practice #4: integral of (x^3)(sqrt(x^2 + 1)) (MIT integration bee 2016 qualifying round)

Hi guys! Here is my integral practice #4. Give it a try first and check the solution in the end. For integral problem requests, just comment down below. Thank you for watching! #integral #calculus

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[felipeleivax]

i did it using trigonometric substitution, using x = tan(θ) and sqrt(x^2+1) = sec(θ). Then you have to integrate tan^3(θ)*sec^3(θ) and using tan^2(θ) = sec^2(θ) -1, you get the integral tan(θ)sec^5(θ) - integral tan(θ)sec^3(θ), which result after using the initial substitution is the same that the video

[lordcezar]

Integration by parts works when you write it as x².x√x²+1

[protocolwonder]

i couldn't hear any sound or explanation

[JesusMartinez-ooht]

Where do you get 1/2 (2 u3/2) (u/5 - 1/3) + c?

[gelbkehlchen]

Solution:
∫x³/√(x²+1)*dx = 1/2*∫x²/√(x²+1)*2*x*dx =
————————————————
Substitution: u = x²+1 x² = u-1 du = 2*x*dx
————————————————
= 1/2*∫(u-1)/√u*du = 1/2*[∫u/√u*du-∫1/√u*du]
= = 1/2*[2/3*u^(3/2)-2*u^(1/2)]+C
= 1/3*u^(3/2)-u^(1/2)]+C =

Checking the result by deriving:

=
= √(x²+1)*x-x/√(x²+1)
= [(x²+1)*x-x]/√(x²+1)
= [x³+x-x]/√(x²+1)
= x³/√(x²+1) everything okay!

[SHUBHAMKRYADAV-wjcn]

Bro I am from India, and I think, this method is very different