A calculus problem by Md Zuhair

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

$\large \displaystyle \frac{dI(k)}{dk} = \int_0^1 \frac{x^k \cdot \ln(x)}{\ln(x)} dx$

$\large \displaystyle \frac{dI(k)}{dk} = \int_0^1 x^k dx$

$\large \displaystyle \frac{dI(k)}{dk} = \frac{1}{k+1}$

$\color{#20A900} \boxed{\large \displaystyle I(k) = \ln(k+1) + C}$

Where $C$ is the arbitrary constant of integration. When $k=0$ :

$\large \displaystyle I(0) = \int_0^1 \frac{x^0 - 1}{\ln(x)} dx$

$\color{#20A900} \boxed{ \large \displaystyle I(0) = 0 }$

So:

$\large \displaystyle I(0) = \ln(1) + C = 0$

$\color{#20A900} \boxed{ \large \displaystyle C = 0 }$

Thus:

$\color{#20A900} \boxed{\large \displaystyle I(k) = \ln(k+1)}$

For $k=201$ :

$\color{#20A900} \boxed{\large \displaystyle I(201) = \int_0^1 \frac{x^{201} - 1}{\ln(x)} dx= \ln(202)}$ .

Thus:

$\color{#3D99F6} \boxed{\large \displaystyle a = 202}$

by the substitution $u = \ln x$ , we get

$\displaystyle \int_0^1\frac{x^k -1}{\ln x}\mathrm dx = \int_0^\infty \frac{1-e^{kx}}{x}e^x\mathrm dx$

which is a Frullani integral

for a function $f: \mathbb{R}\rightarrow \mathbb{R}$ over $x\geq0$ and $f(\infty)$ exists

$\displaystyle \int_0^\infty \frac{f(ax)-f(bx)}{x}\mathrm dx = (f(\infty)-f(0))\ln\frac{a}{b}$

plugging in $a = -1$ , $b=-(n+1)$ , and $f(x) = e^{-x}$ ,

$\displaystyle \int_0^\infty \frac{e^x-e^{(k+1)x}}{u}\mathrm dx = (e^{-\infty}-e^0)\ln(\frac{1}{n+1})=\ln(n+1)$

and therefore $\displaystyle \int_0^1\frac{x^{201} -1}{\ln x}\mathrm dx= \boxed{\ln(202)}$

See @Guilherme Niedu 's nice solution.

You practice more of these types of integrals on the differentiating through the integral page.