Question

Integral of arctan(1/x).

The (1/x) confuses me.

I used integration by parts

u = arctan(1/x), du= 1/(1+(1/x^2))
v=x, dv = 1

I see that my du is wrong, because of reverse chain rule or something, that's the part I don't get.

Can someone explain to me what du would be please?

Thank you

Answer 1

Not reverse chain rule. Just straight chain rule.

du= 1/(1+(1/x^2)) * d/dx (1/x)

Answer 2

\[\text{ let } u=\arctan(\frac{1}{x}) \\
\text{ then } \tan(u)=\frac{1}{x}\\
\text{ so } sec^2(u) du=\frac{-1}{x^2} dx \\
-x^2 sec^2(u) du=dx\\\]
\[-\cot^2(u)\sec^2(u)du=dx \\ -\csc^2(u) du =dx \]
\[\int\limits_{}^{}\arctan(\frac{1}{x})dx=\int\limits_{}^{}-u \csc^2(u)du\]
is probably the way I would go about it
and you know then the integration by parts happens after this

Answer 3

but you can go about it the way you have suggested

Answer 4

@aum gah I feel so silly! I was just complicated it for myself.
Thank you both