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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