How do you integrate #Ln(1+x^2)#?
1 Answer
Explanation:
First, applying integration by parts, we let
Applying the formula
To solve the remaining integral, we will use trig substitution.
Let
Substituting, we have
#=inttan^2(theta)/sec^2(theta)sec^2(theta)d theta#
#=inttan^2(theta)d theta#
#=int(sec^2(theta)-1)d theta#
#=intsec^2(theta)d theta - intd theta#
#=tan(theta)-theta + C#
#=x - arctan(x) + C#
Going back to our original problem, we have
#=xln(1+x^2)-2(x-arctan(x))+C#
#=xln(1+x^2)-2x+arctan(x)+C#