What is the integral of #x^3/(x^2+1)#?

1 Answer
sente
Oct 9, 2016

#intx^3/(x^2+1)dx =(x^2-ln(x^2+1))/2+C#

Explanation:

We will use integration by substitution, as well as the integrals #int1/xdx = ln|x|+C# and #int1dx = x+C#

#intx^3/(x^2+1)dx = intx^2/(x^2+1)xdx#

#=1/2int((x^2+1)-1)/(x^2+1)2xdx#

Let #u = x^2 + 1 => du = 2xdx#. Then

#1/2int((x^2+1)-1)/(x^2+1)2xdx = 1/2int(u-1)/udu#

#=1/2int(1-1/u)du#

#=1/2(u-ln|u|)+C#

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

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

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

(Note that as #C# is an arbitrary constant, we can disregard the #1/2# as we did in the last step. Adding an additional constant makes no difference when we already are considering all functions of that form which vary by a constant.)

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