Is Integration by parts the way to go?

We know that,

$$ $\displaystyle \int_{0}^{1}x^{a}dx = \dfrac{1}{a+1}$

$$ Differentiating under the integral sign with respect to a,

$$ $\displaystyle \int_{0}^{1}x^{a}log_{e}(x)dx = \dfrac{-1}{(a+1)^2}$

$$ Differentiating again with respect to a,

$$ $\displaystyle \int_{0}^{1}x^{a}log_{e}^{2}(x)dx = \dfrac{2}{(a+1)^3}$

$$

Repeating the process,

$$ $\displaystyle \int_{0}^{1}x^{a}log_{e}^{3}(x)dx = \dfrac{-6}{(a+1)^4}$

$$ We notice a pattern that,

$\displaystyle \int_{0}^{1}x^{a}log_{e}^{n}(x)dx = \dfrac{(-1)^{n} \times n!}{(a+1)^{n+1}}$

$$ For the given question, put n = $150$ , a = $50$ ,

$$

$\therefore$ I = $\displaystyle \int_{0}^{1}x^{50}log_{e}^{150}dx = \dfrac{(-1)^{150} \times 150!}{(51)^{151 }}= \dfrac{150!}{51^{151}}$

$$ Comparing, we get, A = 150, B = 51 C = 151

$$

$\therefore A + B + C = 352$

Alternate Solution:

Substitute $\large x=e^{\frac{-t}{51}}$

Then the integral becomes:

$\large \frac { 1 }{ { 51 }^{ 151 } } \int _{ 0 }^{ \infty }{ { e }^{ -t } } { t }^{ 150 }dt\\ \large =\frac { \Gamma \left( 151 \right) }{ { 51 }^{ 151 } } \\ \large =\frac { 150! }{ { 51 }^{ 151 } }$

Let

$\begin{aligned} f(m,n) &= \int_0 ^ 1 x^m(\ln x)^n \\ &= \frac{1}{m+1}x^{m+1}(\ln x)^n \Bigg | ^1 _0 - \frac{n}{m+1}\int_0 ^1 x^m(\ln x)^{n-1} \quad \text{(Integration by Parts)}\\ &= -\frac{n}{m+1} f(m,n-1) \\ &= (-1)^n \frac{n}{m+1} \cdot \frac{n-1}{m+1} \cdot \frac{n-2}{m+1} \cdot \dots \cdot \frac{1}{m+1} \cdot f(m,0) \\ &= (-1)^n \frac{n}{m+1} \cdot \frac{n-1}{m+1} \cdot \frac{n-2}{m+1} \cdot \dots \cdot \frac{1}{m+1} \cdot \frac{1}{m+1} \\ &= \frac{(-1)^n\quad n!}{(m+1)^{n+1}} \end{aligned}$

Notice the problem. The integral contains $\ln x$ and $x$ . If we put $x = e^{-t}$ , we get $-t$ and $e^{-t}$ . Terms present in Gamma. Thus let's get going. Introduction, $\Gamma(n+1) = \int_0^\infty e^{-t}t^{n}\,dt\\\Gamma(n\in\mathbb Z^+) = (n-1)!$

Substitute $x = e^{-t}$ , $dx = -e^{-t} ~dt$ and $\ln x = -t$ $I = \int_0^1 x^m (\ln x)^n \,dx$

Putting in the substitution, $I = \int_0^\infty e^{-(m+1)t}\, (-t)^n \, dt$

Again substituting $(m+1)t = x$ , $\dfrac{dx}{m+1} = dt$ . So, $\begin{aligned}I &= \int_0^\infty e^{-x} \left(\dfrac{-x}{m+1}\right)^n \dfrac{dx}{m+1}\\ &=\dfrac{(-1)^n}{(m+1)^{n+1}} \int_0^\infty e^{-x} x^n \, dx\\&= \dfrac{(-1)^n\Gamma(n+1)}{(m+1)^{n+1}} \end{aligned}$

$\boxed{\displaystyle\int_0^1 x^m (\ln x)^n \,dx = \dfrac{(-1)^n\cdot\Gamma(n+1)}{(m+1)^{n+1}}}$

Therefore, if $n\in \mathbb Z^+$ , $\int_0^1 x^m (\ln x)^n \,dx = \dfrac{(-1)^n\cdot n!}{(m+1)^{n+1}}$