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)

Question

anonymous · Answer

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

anonymous · Answer

hi

anonymous · Answer

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

let $$u = 1+\arctan(x)$$

then $$du/dx = 1/(x^2+1)$$

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

anonymous · Answer

thank you! I