<s>[INST] Prove that: $\frac1{20}\le \int_{1}^{\sqrt 2} \frac{\ln x}{\ln^2x+1} dx$ I'm interested in proving the following integral inequality:
$$\frac1{20}\le \int_{1}^{\sqrt 2} \frac{\ln x}{\ln^2x+1} dx$$
According to W|A the result of this integral isn't pretty nice, and involves the exponential integral.
 [/INST] $$\begin{array}{c l}  \int_1^{\sqrt2} \frac{\log x}{1+(\log x)^2}dx & =\int_0^{\frac{1}{2}\log2}\frac{u}{1+u^2}e^udu \\ & \ge\int_0^{\frac{1}{2}\log2}\frac{u}{1+u^2}du \\ & = \left[\frac{1}{2}\log(1+u^2)\right]_0^{\frac{1}{2}\log 2} \\ & =\frac{1}{2}\log\left(1+\frac{(\log 2)^2}{4}\right) \\ & \approx 0.056714899 \\ & > 0.05=\frac{1}{20}. \end{array}$$

Edit. We can make it tighter with $e^u\ge1+u$:
$$\begin{array}{c l}  \int_1^{\sqrt2} \frac{\log x}{1+(\log x)^2}dx & =\int_0^{\frac{1}{2}\log2}\frac{u}{1+u^2}e^udu \\ 
& \ge\int_0^{\frac{1}{2}\log2}\frac{u}{1+u^2}(1+u)du \\
& = \int_0^{\frac{1}{2}\log2}\left(1+\frac{u}{1+u^2}-\frac{1}{1+u^2}\right)du \\
& = \left[u+\frac{1}{2}\log(1+u^2)-\arctan u\right]_0^{\frac{1}{2}\log 2} \\ 
& =\frac{1}{2}\log\left(2+\frac{(\log 2)^2}{2}\right)-\arctan\left(\frac{1}{2}\log 2\right) \\ 
& \approx 0.06\color{Blue}{96694}. \end{array}$$
</s>