Find the area bounded between $$f(x)=\frac{\arctan(x)}{x^2} \quad\text{and}\quad g(x)=\frac{\arctan(x)}{x^2+1}.$$
The title says the question. The limits are from 1 to infinity. I know that I must do this: $$\int_1^\infty [f(x)-g(x)]\,dx$$
The integral of g(x) is trivial, it is a simple u-substitution.But for the integral of $f(x)$, I must do integration by parts. I set $u=arctan(x)$ and $dv=1/x^2$. The problem lies in the integral of udv:
$$\int^\infty_1\frac{1}{x(x^2+1)}dx$$ When we do the partial fraction decomp. we get:
$$\int^\infty_1\left(\frac{1}{x} - \frac{x}{x^2+1}\right)dx$$
Now, we are going to get two natural logs, $$\ln(x), \quad \frac{1}{2} \ln(x^2+1).$$
Since this is going to infinity, won't the logs also go to infinity, leaving us with the volume being infinite? What am I doing wrong?
As $x \to \infty$, $$\log \left( \frac{x}{\sqrt{x^2+1}} \right) \to 0.$$