Indefinite Integrals — Question 5

Question 5 - Solution

Split the integral: $$\int \left( \frac{1}{x}\arctan x + \frac{x^2}{1+x^2} \right)\,dx = \int \frac{\arctan x}{x}\,dx + \int \frac{x^2}{1+x^2}\,dx.$$

1) The rational part

Rewrite: $$\frac{x^2}{1+x^2}=1-\frac{1}{1+x^2}.$$ So $$\int \frac{x^2}{1+x^2}\,dx = \int\left(1-\frac{1}{1+x^2}\right)\,dx = x-\arctan x.$$

2) The non-elementary part $\displaystyle \int \frac{\arctan x}{x}\,dx$

A common attempt is integration by parts, which produces $$\int \frac{\arctan x}{x}\,dx = \arctan x\,\ln|x| - \int \frac{\ln|x|}{1+x^2}\,dx,$$ and then the substitution $x=\tan\theta$ turns the remaining integral into $$\int \frac{\ln|x|}{1+x^2}\,dx = \int \ln(\tan\theta)\,d\theta.$$ The key clarification is:

$\displaystyle \int \ln(\tan\theta)\,d\theta$ is not an elementary antiderivative. So any step claiming $$\int \ln(\tan\theta)\,d\theta = \theta\ln(\tan\theta)-\tan\theta + C$$ is incorrect (differentiate it to check).

A correct closed form uses the dilogarithm function $\operatorname{Li}_2$. Start from $$\arctan x=\frac{1}{2i}\left(\ln(1+ix)-\ln(1-ix)\right).$$ Then $$\int \frac{\arctan x}{x}\,dx = \frac{1}{2i}\int \frac{\ln(1+ix)-\ln(1-ix)}{x}\,dx.$$ Use the identity $$\frac{d}{dx}\operatorname{Li}_2(u)=-\ln(1-u)\,\frac{u'}{u},$$ so, for real $x$, $$\frac{d}{dx}\operatorname{Li}_2(-ix)=-\frac{\ln(1+ix)}{x}, \qquad \frac{d}{dx}\operatorname{Li}_2(ix)=-\frac{\ln(1-ix)}{x}.$$ Therefore, $$\int \frac{\ln(1+ix)}{x}\,dx=-\operatorname{Li}_2(-ix)+C, \qquad \int \frac{\ln(1-ix)}{x}\,dx=-\operatorname{Li}_2(ix)+C,$$ and hence $$\int \frac{\arctan x}{x}\,dx = \frac{1}{2i}\Bigl(\operatorname{Li}_2(ix)-\operatorname{Li}_2(-ix)\Bigr)+C.$$

Final Answer

Combine both parts: $$\int \left( \frac{1}{x}\arctan x + \frac{x^2}{1+x^2} \right)\,dx = \left(x-\arctan x\right) + \frac{1}{2i}\Bigl(\operatorname{Li}_2(ix)-\operatorname{Li}_2(-ix)\Bigr) + C.$$