AP Calculus AB and BC

Integral of x^3/(x^2+1): Divide First

The integral of x cubed over x squared plus 1 is x squared over 2, minus the natural log of x squared plus 1 over 2, plus C. The numerator degree exceeds the denominator's, so divide first: the fraction becomes x minus x over x squared plus 1.

∫x3x2+1 dx=x22−ln⁡(x2+1)2+C\int \frac{x^{3}}{x^{2}+1}\,dx = \frac{x^{2}}{2} - \frac{\ln\left(x^{2}+1\right)}{2} + C

Long division, then substitution

x3x2+1=x−xx2+1\frac{x^{3}}{x^{2}+1} = x - \frac{x}{x^{2}+1}

The first piece integrates by the power rule and the second by the substitution u=x2+1u = x^{2}+1, which gives half a logarithm.

∫x3x2+1dx=x22−12ln⁡(x2+1)+C\int \frac{x^{3}}{x^{2}+1}dx = \frac{x^{2}}{2} - \frac{1}{2}\ln\left(x^{2}+1\right) + C

Spotting an improper fraction

If the numerator's degree is at least the denominator's, no substitution will finish the job on its own. Dividing is Topic 6.10, and recognising the need for it is most of the work.

Common mistakes

  • Substituting u=x2+1u = x^{2}+1 straight away, which leaves an x2x^{2} that has to be rewritten as u−1u-1 anyway.
  • Producing an arctangent. That needs a CONSTANT numerator.

Every answer on this page is machine checked

An automated test differentiates the antiderivative above and confirms it returns the integrand. A wrong sign or a missing factor fails the build, so it cannot reach you.

Frequently asked questions

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

It is x22−12ln⁡(x2+1)+C\frac{x^{2}}{2} - \frac{1}{2}\ln\left(x^{2}+1\right) + C.

Why divide first?

The fraction is improper, so no substitution finishes it directly.