<s>[INST] How to evaluate the integral $\int_{0}^{1} \frac{\log x}{\sqrt {1+x^2}}dx$ $$I=\int_{0}^{1} \frac{\log x}{\sqrt {1+x^2}}dx$$
My attempt:$$I=\int_{0}^{1}\log x d(\log(x+\sqrt{1+x^2}))$$
$$=\log x\log(x+\sqrt{1+x^2})|_0^1-\int_{0}^{1}\frac{\log(x+\sqrt{1+x^2})}{x}dx$$
I don't know how to proceed below, please help me.
That is different to me.
 [/INST] \begin{align}
\int_0^1 x^\alpha\ dx &= \dfrac{x^{\alpha+1}}{\alpha+1}\\
\dfrac{d}{d\alpha}\int_0^1 x^\alpha\ dx &=\dfrac{d}{d\alpha}\dfrac{x^{\alpha+1}}{\alpha+1}\\
\int_0^1 x^\alpha\ln x\ dx &=\dfrac{x^{\alpha+1}((\alpha+1)\ln x-1)}{(\alpha+1)^2}\Big|_0^1=\dfrac{-1}{(\alpha+1)^2}
\end{align}
\begin{align}
I
&= \int_{0}^{1} \frac{\ln x}{\sqrt {1+x^2}}dx \\
&= \int_0^1\ln x\left(1-\dfrac12x^2+\dfrac38x^4+\cdots\right)\ dx\\
&= \int_{0}^{1} \frac{\ln x}{\sqrt {1+x^2}}dx \\
&= \int_0^1\left(\ln x-\dfrac12x^2\ln x+\dfrac38x^4\ln x+\cdots\right)\ dx \\
&= -1+\dfrac{1}{18}-\dfrac{3}{200}+\cdots
\end{align}
</s>