Question

integral arctan sqrt x/(sqrt x(1+x))

Answer 1

i know \[u=\arctan \sqrt{x} \]

Answer 2

and \[du= 1/ \sqrt{x} (2x+2) dx\]

Answer 3

do u mean
\[\large \int\frac{\arctan\sqrt{x}}{\sqrt{x+x^2}}\,dx \]
?

Answer 4

no, i mean \[\int\limits\limits \arctan \sqrt{x} / \sqrt{x} (1+x) dx\]

Answer 5

\[\int\limits \frac{ \arctan \sqrt{x}}{ \sqrt{x}(1+x) } dx\]

Answer 6

u were right
\[\large u=\arctan\sqrt{x} \]
so
\[\large du=\frac{1}{1+x}\cdot\frac{1}{2\sqrt{x}}\,dx \]
so
\[\large 2du=\frac{dx}{\sqrt{x}(1+x)} \]
so the integral becomes
\[\large \int u\,2du=2\frac{u^2}{2}=u^2=\arctan^2\sqrt{x}+C \]

Answer 7

thanks!

Answer 8

u r welcome