Question

can someone show me how to integrate (9(1+arctan x))/(1+x^2). or can you tell me the table number i should use (stewart book)

Answer 1

if I substitue u = arctan x, du = 1/(1+x^2), dx = 1+x^2 du this would cancel 1+x^2 at the bottom of the function and I'm stuck integrating 9(1+U) du

Answer 2

hi

Answer 3

\[\int\limits 9(1+\arctan(x)) / (1+x^2)\]

let \[u = 1+\arctan(x)\]

then \[du/dx = 1/(x^2+1)\]

so \[\int\limits 9(1+\arctan(x))/(1+x^2) dx = \int\limits (9u)du \]
\[=9u^2/2 +c\]
\[= 9(1+\arctan(x))^2 /2 +c\]
\[=9(\arctan(x)) +(9/2)(\arctan(u)^2 +c_2\]

Answer 4

thank you! I