$J=\int_{0}^{1}{ \frac{x^b-x}{\ln x}}dx$ We know how to solve:$$I=\int_{0}^{1}{ \frac{x^b-1}{\ln x}}dx$$
Let $$f(b)=\int_{0}^{1} \dfrac{x^b-1}{\ln x} dx$$
$$f'(b)=\int_{0}^{1} x^b dx$$
$$f'(b)=\dfrac{1}{b+1}$$
$$f(b)=\ln(b+1)+C$$
Let $b=0$ then $f(b)=0$ implies $C=0 $
Therefore $f(b)=\ln(b+1)$
My question: 
How evaluating the following integral?

$$J=\int_{0}^{1}{ \frac{x^b-x}{\ln x}}dx$$

 A: Concerning
$$
J=\int_{0}^{1}{ \frac{x^b-x}{\ln x}}dx
$$ one may just write
$$
J=\int_{0}^{1}{ \frac{(x^b-1)-(x-1)}{\ln x}}dx=\int_{0}^{1}{ \frac{x^b-1}{\ln x}}dx-\int_{0}^{1}{ \frac{x-1}{\ln x}}dx
$$  then one may use the previous result.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\int_{0}^{1}{x^{b} - 1 \over \ln\pars{x}}\,\dd x & =
\int_{0}^{1}\pars{1 - x^{b}}\int_{0}^{\infty}x^{t}\,\dd t\,\dd x =
\int_{0}^{\infty}\int_{0}^{1}\pars{x^{t} - x^{t + b}}\,\dd x\,\dd t
\\[5mm] & =
\int_{0}^{\infty}\pars{{1 \over t + 1} - {1 \over t + b + 1}}\,\dd t =
\left.\ln\pars{t + 1 \over t + b + 1}\right\vert_{\ t\ =\ 0}^{\ t\ \to\ \infty} =
\bbx{\ds{\ln\pars{b + 1}}}
\end{align}
