So I'm trying to solve for the indefinite integral $$\int \frac{\arctan(x)}{x^3}\, dx$$
Here is what I tried:
Use integration by parts:
$$\int v\,du = vu – \int u\,dv$$
Let $v = \arctan(x)\\$
Let $dv = \frac{dx}{1 + x^2}.\\$
Let $du = x^3\, dx.\\$
Let $u = -\frac{1}{2x^2}.\\$
$$ -\frac{\arctan(x)}{2x^2} + \frac{1}{2}\int \frac{1}{x^2} \cdot \frac{1}{1 + x^2} \,dx$$
Now I'm sort of stuck, and since we haven't been taught partial fractions, I'm not sure how to proceed. Would someone mind explaining how to approach this? Thanks in advance!
Best Answer
Hint. One may just write $$ \frac{1}{x^2(1 + x^2)}=\frac{(1+x^2)-x^2}{x^2(1 + x^2)}=\frac{1}{x^2}-\frac{1}{1 + x^2}. $$