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

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


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.
