What is the definite integral of arctan(1/x) evaluated from 1 to sqrt(3)?

Question

anonymous · Answer

Integration by parts seems to be the way to go as arctan will become nicer upon differentiation

anonymous · Answer

satellite73  please help me

anonymous · Answer

$$\begin{matrix}u=\arctan\left(\frac{1}{x}\right)&dv=dx\
du=\frac{-\frac{1}{x^2}}{\frac{1}{x^2}+1}dx&v=x\end{matrix}$$
$$x\arctan\left(\frac{1}{x}\right)-\int x\frac{-\frac{1}{x^2}}{\frac{1}{x^2}+1}dx\
x\arctan\left(\frac{1}{x}\right)+\int \frac{\frac{1}{x}}{\frac{1}{x^2}+1}dx$$
This integral seems tricky at first... Not sure if there's a closed form.

anonymous · Answer

$$x\arctan\left(\frac{1}{x}\right)-\int\frac{1}{\left(\frac{1}{x^2}+1\right)x}dx$$
That might help a bit.

anonymous · Answer

okk thanks :)

zepdrix · Answer

Multiplying that last integral $\large \int \dfrac{\frac{1}{x}}{\frac{1}{x^2}+1}dx$ by $\large \dfrac{x^2}{x^2}$ should clean it up a little bit. :)