I looked up how to do this again a few days ago. I'm trying to remember this too.
Integration by parts of \(tan^{-1}(x) \times 1\).
\(tan^{-1}(x) \times \int (1)dx - \int (\frac{d}{dx}tan^{-1}(x) \times \int (1)dx) dx\)
\(tan^{-1}(\theta) \times x - \int (\frac{1}{1+x^2} \times x)dx\)
\(xtan^{-1}(x) - \int (\frac{x}{1+x^2})dx \)
Substituting \(u=1+x^2\)
\(du=2x\ dx\)
\(dx=\frac{du}{2x}\)
\(xtan^{-1}(x) - \int (\frac{x}{u} \times \frac{du}{2x})\)
\(xtan^{-1}(x) - \int (\frac{1}{2u} ) du\)
\(xtan^{-1}(x) - \frac{1}{2} \int (\frac{1}{u}) du\)
\(xtan^{-1}(x) - \frac{1}{2} [ln(u)]\)
\(u=1+x^2\), hence:
\(xtan^{-1}(x) - \frac{1}{2}ln(1+x^2)\)
There's our integral!
This article was written on 30/01/2023. If you have any thoughts, feel free to send me an email with them. Have a nice day!