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)
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
\[\int\limits 9(1+\arctan(x)) / (1+x^2)\] let \[u = 1+\arctan(x)\] then \[du/dx = 1/(x^2+1)\] so \[\int\limits 9(1+\arctan(x))/(1+x^2) dx = \int\limits (9u)du \] \[=9u^2/2 +c\] \[= 9(1+\arctan(x))^2 /2 +c\] \[=9(\arctan(x)) +(9/2)(\arctan(u)^2 +c_2\]
thank you! I
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