Indefinite Integrals — Question 9

Question 9

Find the general antiderivative of $$\int \frac{\arctan x}{x^2}\,dx.$$

Question 9 - Solution

We use integration by parts.

Let $$u=\arctan x, \qquad dv=\frac{1}{x^2}\,dx.$$ Then $$du=\frac{1}{1+x^2}\,dx, \qquad v=-\frac{1}{x}.$$

Applying integration by parts, $$\int \frac{\arctan x}{x^2}\,dx = -\frac{\arctan x}{x} + \int \frac{1}{x(1+x^2)}\,dx.$$

Now decompose the integrand: $$\frac{1}{x(1+x^2)} = \frac{1}{x} - \frac{x}{1+x^2}.$$

Thus, $$\int \frac{1}{x(1+x^2)}\,dx = \ln|x| - \frac{1}{2}\ln(1+x^2).$$

Final Answer: $$\boxed{ -\frac{\arctan x}{x} + \ln|x| - \frac{1}{2}\ln(1+x^2) + C }$$