AP Calculus AB and BC

Integral of x^2/(x^3+1): Answer and Steps

Q: Why does the answer have 1/3 in front?

Because du = 3x^2\,dx while the integral offers only x^2\,dx, which is \frac{du}{3}. That \frac{1}{3} rides along to the end.

Q: Can I drop the absolute value if x is positive?

On an interval where x^3+1 > 0 you may write \frac{1}{3}\ln(x^3+1) + C and nothing is lost. A general antiderivative comes with no stated domain, so keep the bars unless the problem pins x down.

Q: Does the same trick work for x/(x^3+1)?

No. x is not a constant multiple of 3x^2, so the pattern fails. That integral needs partial fractions after factoring x^3+1 = (x+1)(x^2-x+1), which goes past what AP asks for.

The integral of x^2/(x^3 + 1) is (1/3)ln|x^3 + 1| + C. The numerator is one third of the derivative of the denominator, so u = x^3 + 1 turns the problem into (1/3) times the integral of du/u. The absolute value keeps the answer valid where x^3 + 1 is negative.

\int \frac{x^2}{x^3+1}\,dx = \frac{1}{3}\ln|x^3+1| + C

Spotting the derivative in the numerator

The denominator $x^3+1$ has derivative $3x^2$ , and the numerator is $x^2$ , exactly one third of it. Any fraction with that shape integrates to a logarithm.

\int \frac{f'(x)}{f(x)}\,dx = \ln|f(x)| + C

Make it explicit with $u = x^3+1$ , so $du = 3x^2\,dx$ and $x^2\,dx = \frac{du}{3}$ . The mismatch between $x^2$ and $3x^2$ is the only reason a constant appears out front.

\int \frac{x^2}{x^3+1}\,dx = \int \frac{1}{u}\cdot\frac{du}{3} = \frac{1}{3}\int \frac{du}{u}

Back-substituting, with the bars

\frac{1}{3}\int \frac{du}{u} = \frac{1}{3}\ln|u| + C = \frac{1}{3}\ln|x^3+1| + C

The bars are load bearing. For $x < -1$ the quantity $x^3+1$ is negative: at $x = -3$ the integrand is $\frac{9}{-26}$ , a perfectly ordinary number, while $\ln(x^3+1)$ would be undefined there. Absolute value lets one expression cover both sides.

Differentiating confirms it. The chain rule on $\ln|x^3+1|$ gives $\frac{3x^2}{x^3+1}$ on either branch, and the $\frac{1}{3}$ trims it back to the integrand.

Where the antiderivative is valid

The integrand blows up at $x = -1$ , so an antiderivative only describes the area on an interval that avoids that point. A definite integral whose limits straddle $x = -1$ diverges no matter how the bars are written.

The mistake students make

The first error is losing the $\frac{1}{3}$ and writing $\ln|x^3+1| + C$ . Differentiating that returns $\frac{3x^2}{x^3+1}$ , three times the integrand. The constant is there because $du$ demanded $3x^2\,dx$ and the problem only supplied $x^2\,dx$ .

The second is splitting the fraction, as in $\frac{x^2}{x^3} + \frac{x^2}{1}$ . Only a sum in the numerator can be split across a common denominator, never a sum in the denominator, and the split version is a different function.

Test for the log pattern first

Before reaching for anything harder, differentiate the denominator and compare with the numerator. If they match up to a constant multiple, the answer is that constant times $\ln$ of the absolute value of the denominator.

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

Why does the answer have 1/3 in front?

Because $du = 3x^2\,dx$ while the integral offers only $x^2\,dx$ , which is $\frac{du}{3}$ . That $\frac{1}{3}$ rides along to the end.

Can I drop the absolute value if x is positive?

On an interval where $x^3+1 > 0$ you may write $\frac{1}{3}\ln(x^3+1) + C$ and nothing is lost. A general antiderivative comes with no stated domain, so keep the bars unless the problem pins $x$ down.

Does the same trick work for x/(x^3+1)?

No. $x$ is not a constant multiple of $3x^2$ , so the pattern fails. That integral needs partial fractions after factoring $x^3+1 = (x+1)(x^2-x+1)$ , which goes past what AP asks for.