Integral ( x^5 e^x^4 dx from 0 to infinity

[@MathFromAlphaToOmega]

A little shortcut if you're familiar with the gamma function is to let x=u^(1/4), and then the integrand becomes 1/4*sqrt(u)e^(-u). That evaluates to 1/4*Γ(3/2)=1/8*Γ(1/2), which is sqrt(pi)/8.

[@adw1z]

Nice, as soon as you see the limits 0 and ∞ and the integrand is of the form x^m exp(x^n), the gamma function immediately springs to mind

[@blz2151]

​@@adw1zaà

[@sriprasadjoshi3036]

Last lines were fabulously inspiring, thank you...

[@AubreyForever]

The answer is incredible.

[@38rohanjadhav55]

Sir will you make a video on Integral (sec^3(x))

[@sajuvasu]

What if we interchange the 4 and 5 😂

[@nanamacapagal8342]

So much easier!

[@Subham-Kun]

You are pure evil 👿

[@necrolord1920]

4:32 should be te^(-t^2) dt. You left out the differential.

[@Brid727]

sir, IBP will be much faster by the DI tabular method(from bprp) so I suggest you use that

[@Orillians]

Can you send a video linking his tabular method ( I have seen his tabular method but Idk how to use it)

[@necrolord1920]

Tabular integration by parts doesn't save much time if you are only using integration by parts once, which is the case here.

[@Brid727]

@@necrolord1920my idea is that if we let t be the u, differentiating it twice will give 0 and for that we’ll need to integrate the rest of the integrand twice but trust me it’ll work out and if we do that, we can get rid of the final integral we would otherwise need to do in normal IBP and have a full answer in one take though yeah we’d end up with the answer prime newtons got, we would save a bit of time from not writing the uv- integral of v du