How to integrate #int 1/(1+(x+2)^2)dx# ?

1 Answer
Jun 23, 2018

Shown below

Explanation:

let # u = x+2 #

#du = dx #

#=> int 1/(1+u^2) du #

# u = tan theta #

#du = sec^2 theta d theta #

#=> int (sec^2 theta d theta ) /(1+tan^2 theta ) #

#=> int (sec^2 theta d theta ) / sec^2 theta #

#=> int 1 d theta #

#=> theta + c #

#=> arctan u + c #

#=color(red)( arctan(x+2) + c #