Question

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

Answer 1

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

Answer 2

satellite73 please help me

Answer 3

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

Answer 4

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

Answer 5

okk thanks :)

Answer 6

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. :)