<s>[INST] evaluate $\int^{\infty}_1\frac{\ln(x)}{1+x^2}dx$ I need to calculate the integral: $$\int^{\infty}_1\frac{\ln(x)}{1+x^2}dx$$
My attempt:
IBP: $u=\ln(x) \Rightarrow du=\frac{1}{x}dx, vdv=\frac{1}{1+x^2}dx \Rightarrow v=\arctan(x)$
So, the integral becomes:
$$\ln(x)\arctan(x)\Big|^{\infty}_1-\int^{\infty}_1\frac{\arctan(x)}{x}dx=\ln(x)\arctan(x)\Big|^{\infty}_1-x+\frac{x^3}{3^2}-\frac{x^5}{5^2}+\frac{x^7}{7^2}-...\Big|^{\infty}_1$$
Since the integral $$\int^{\infty}_1\frac{\arctan(x)}{x}dx=\int^{\infty}_1\sum^{\infty}_{n=0}(-1)^n\frac{x^{2n+1}}{(2n+1)x}dx=\int^{\infty}_1\sum^{\infty}_{n=0}\frac{x^{2n}}{2n+1}dx=\sum^{\infty}_{n=0}(-1)^n\frac{x^{2n+1}}{(2n+1)^2}\Big|^{\infty}_1=x-\frac{x^3}{3^2}+\frac{x^5}{5^2}-\frac{x^7}{7^2}+...\Big|^{\infty}_1$$
Is this correct?
Also, this may help with the integral with lower bound $0$
https://www.youtube.com/watch?v=JCZFVQ9uFHg
 [/INST] Convert the limits to $(0,1)$ to have the expansion converging,
$$\begin{align}
&\int^{\infty}_1\frac{\ln x}{1+x^2}dx
=- \int_{0}^1\frac{\ln t }{1+t^2}dt\\
& =-\ln t\arctan t\Big|_0^1+\int^{1}_0\frac{\arctan t}{t}dt \\
&=\int^{1}_0\frac1t (t-\frac{t^3}3 + \frac{t^5}5 - ...)dt 
=(1-\frac{t^3}{3^2}+ \frac{t^5}{5^2}- ...)\bigg|^{1}_0\\
&= 1- \frac1{3^2} + \frac1{5^2}  -\frac1{7^2}  +...
\end{align}$$
</s>