Integration -3

$\large \displaystyle I(n) = \int_0^1 \frac{ \ln(x)^n}{\sqrt{x}} dx$

$\color{#20A900} \large \displaystyle \ln(x) = u \rightarrow dx = x \cdot du = e^u \cdot du$

$\large \displaystyle I(n) = \int_{-\infty}^0 u^n \cdot e^{\frac{u}{2}} du$

Integrating by parts:

$\large \displaystyle I(n) = 2 \cdot u^n \cdot e^{\frac{u}{2}} \Bigg |_{- \infty}^0 - 2 \cdot \int_{-\infty}^0 nu^{n-1} \cdot e^{\frac{u}{2}}du$

$\large \displaystyle I(n) = - 2 \cdot \int_{-\infty}^0 nu^{n-1} \cdot e^{\frac{u}{2}}du$

Repeat it $n-1$ more times, totalizing $n$ times and you'll get:

$\large \displaystyle I(n) = (-1)^n \cdot n! \cdot 2^n \int_{-\infty}^0 e^{\frac{u}{2}} du$

$\color{#20A900} \boxed {\large \displaystyle I(n) = (-1)^n \cdot n! \cdot 2^{n+1} }$

So, both $(C)$ and $(D)$ are false. Also:

$\color{#20A900} \boxed {\large \displaystyle \frac{I(n)}{I(n-1)} = -2n}$

So, $(B)$ is also false. The only option which is true is $\color{#3D99F6} \boxed{(A)}$

@Indraneel Mukhopadhyaya

the relation i get between I(n) and I(n-1) is that their ratio is -2n . but i am not able to comment on option D and C