How do you find the integral of # x^3/(1+x^2)#?

1 Answer
Nico Ekkart
Oct 10, 2015

#1/2(1+x^2-ln(1+x^2)) + C#

Explanation:

We'll do integration by substitution. Take:
#u = x^2#
#du = 2x#
Substituting this in, we get:

#1/2\intu/(1+u)du#
Now take:
#t = 1 + u#
#dt = du#

Substitute this in:
#1/2\int(t-1)/tdt = 1/2\int1 - 1/tdt = 1/2(\int1dt - \int 1/tdt)#
#=1/2(t - ln(t)) + C#
Now we need to change back to #x#:
#=1/2(1+u-ln(1+u)) + C#
#=1/2(1+x^2-ln(1+x^2)) + C#

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