AP Calculus AB and BC

Integral of x/(x+1): Divide First

The integral of x over x plus 1 is x minus the natural log of the absolute value of x plus 1, plus C. Because the numerator and denominator have the same degree, divide first: the fraction becomes 1 minus 1 over x plus 1.

∫xx+1 dx=x−ln⁡∣x+1∣+C\int \frac{x}{x+1}\,dx = x - \ln\left|x+1\right| + C

Divide before you integrate

A rational function whose numerator degree is at least the denominator's is IMPROPER, and long division is the first move.

xx+1=(x+1)−1x+1=1−1x+1\frac{x}{x+1} = \frac{(x+1)-1}{x+1} = 1 - \frac{1}{x+1}
∫(1−1x+1)dx=x−ln⁡∣x+1∣+C\int \left(1 - \frac{1}{x+1}\right)dx = x - \ln\left|x+1\right| + C

The trick in one line

Add and subtract to build the denominator in the numerator: x becomes (x+1) - 1. That is faster than setting up long division and works for every linear-over-linear case.

Where it appears

Integrating using long division is Topic 6.10, on both AB and BC. Recognising that a substitution alone cannot finish this one is the point: u=x+1u = x+1 leaves an xx in the numerator that still has to be rewritten as u−1u - 1.

Common mistakes

  • Splitting into ∫x dx\int x\,dx over ∫(x+1) dx\int (x+1)\,dx. There is no such rule for integrals of quotients.
  • Answering ln⁡∣x+1∣\ln\left|x+1\right| alone, forgetting the leading xx.
  • Getting the sign wrong. The rewritten form SUBTRACTS 1x+1\frac{1}{x+1}.

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/(x+1)?

It is x−ln⁡∣x+1∣+Cx - \ln\left|x+1\right| + C.

Why divide first?

The fraction is improper, so no substitution finishes it directly. Rewriting as 1−1x+11 - \frac{1}{x+1} makes both pieces basic.