How do you integrate #Ln(1+x^2)#?

1 Answer
Jul 17, 2016

#intln(1+x^2)dx=xln(1+x^2)-2x+arctan(x)+C#

Explanation:

First, applying integration by parts, we let

#u = ln(1+x^2)# and #dv = dx#
#=> du = (2x)/(1+x^2)# and #v = x#

Applying the formula #intudv = uv-intvdu#, we have

#intln(1+x^2)dx = xln(1+x^2)-2intx^2/(1+x^2)dx#


To solve the remaining integral, we will use trig substitution.

Let #x = tan(theta)#
#=> dx = sec^2(theta)d theta# and #theta = arctan(x)#

Substituting, we have

#intx^2/(1+x^2)dx = inttan^2(theta)/(1+tan^2(theta))sec^2(theta)d theta#

#=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

#intln(1+x^2)dx = xln(1+x^2)-2intx^2/(1+x^2)dx#

#=xln(1+x^2)-2(x-arctan(x))+C#

#=xln(1+x^2)-2x+arctan(x)+C#