It is well-known

$$\int_{0}^{1}\left({1\over \ln(x)}+{1\over 1-x}\right)\mathrm dx=\gamma\tag1$$

Messing around with $(1)$ using wolfram integrator, we have

$$\int_{0}^{1}\left({x^u\over \ln(x)}+{x^v\over 1-x}\right)\mathrm dx=F(u,v)\tag2$$


How do we show that $F(u,v)=\gamma-H_v+\ln(u+1)?$


Use \begin{eqnarray*} H_v= \int_0^1 \frac{1-x^v}{1-x} dx \end{eqnarray*} and \begin{eqnarray*} \frac{x^u-1}{\ln x} = \int_0^u x^t dt. \end{eqnarray*} We have \begin{eqnarray*} \int_0^1 ( \frac{x^u}{\ln x}+ \frac{x^v}{1-x}) dx &=& \int_0^1 ( \frac{1}{\ln x}+ \frac{1}{1-x}- \frac{1-x^v}{1-x}+\frac{x^u-1}{\ln x}) dx \end{eqnarray*} The first two terms of the RHS give $ \gamma $ , the third term gives $H_v$ and the fourth term is \begin{eqnarray*} \int_0^1 \frac{x^u-1}{\ln x} dx = \int_0^1 \int_0^u x^t dt dx = \int_0^u \int_0^1 x^t dx dt = \int_0^u \frac{1}{1+t} dt = \ln (1+u). \end{eqnarray*}


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