Solution to $\int\frac{\ln(1+x^2)}{x^2}dx$

Question

Q: $\int\frac{\ln(1+x^2)}{x^2}dx$

Here is my entire working:

So, overall, I started with the reverse product rule, then onto reverse chain rule and then tried to partial fraction, however, I still got the wrong answer. I feel that I went wrong in the partial fraction decomposition part(I set $B = 1$) but am not too sure so I need some guidance. Thanks!

Answer 1

OK, I've decoded your handwriting and the real problem is when you try to integrate $2/(1+x^2)$. That's just an $\arctan x$.

Answer 2

To begin with, notice that \begin{align*} \int\frac{\ln(1 + x^{2})}{x^{2}}\mathrm{d}x = -\frac{\ln(1 + x^{2})}{x} + \int\frac{2x}{x(x^{2} + 1)}\mathrm{d}x \end{align*} where the last integral equals \begin{align*} \int\frac{2x}{x(x^{2} + 1)}\mathrm{d}x = \int\frac{2}{1 + x^{2}}\mathrm{d}x = 2\arctan(x) + c \end{align*}

Gathering all the previous results, one gets that \begin{align*} \int\frac{\ln(1 + x^{2})}{x^{2}}\mathrm{d}x = -\frac{\ln(1 + x^{2})}{x}+ 2\arctan(x) + c \end{align*} and we are done.

Hopefully this helps!