# How apply beta integral to solve?integral between 1 and e (1/x*ln^3x)(1-lnx)^4dx

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### 1 Answer

You need to remember `beta` function, called Euler integral such that:

`beta(a,b) = int_0^1 t^(a-1)*(1-t)^(b-1)dt= (Gamma(a)*Gamma(b))/(Gamma(a+b))`

You need to remember the factorial formula of `Gamma(n)` such that:

`Gamma(n) = (n-1)!`

You need to solve `int_1^e ((1/x)*ln^3 (x))(1-lnx)^4dx` , hence you need to come up with the substitution `ln x = t =gt (dx)/x = dt` .

You need to write the integral using the variable t such that:

`int_0^1 t^3*(1-t)^4 dt = Beta(4,5) = (Gamma(4)*Gamma(5))/(Gamma(4+5)) = (Gamma(4)*Gamma(5))/(Gamma(9)) `

You need to evaluate `Gamma(4), Gamma(5), Gamma(9)` using `Gamma(n) = (n-1)!:`

`` `Gamma(4) = (4-1)! = 3!`

`Gamma(5) = (5-1)! = 4!`

`` `Gamma(9) = (9-1)! = 8!`

Plugging the values of `Gamma(4), Gamma(5), Gamma(9)` in equation of `beta(4,5)` yields:

`beta (4,5) = (3!*4!)/(8!) = (1*2*3)/(5*6*7*8) = 1/280`

**Hence, evaluating the beta integral yields`int_1^e ((1/x)*ln^3 (x))(1-lnx)^4dx= 1/280.` **