AP Calculus BC

Integral of 1/(x^2+x): Partial Fractions to One Log

The integral of 1 over x squared plus x is the natural log of the absolute value of x over x plus 1, plus C. The denominator factors into x times x plus 1, so partial fractions splits the fraction into 1 over x minus 1 over x plus 1, and the quotient law folds the two logs into one.

∫1x2+x dx=ln⁡∣xx+1∣+C\int \frac{1}{x^{2}+x}\,dx = \ln\left|\frac{x}{x+1}\right| + C

Factor, then split

The obvious substitutions go nowhere on 1x2+x\frac{1}{x^{2}+x}: the numerator is not the derivative of the denominator, nor a constant multiple of it, so the direct log pattern does not apply. What it does have is a denominator that factors: x2+x=x(x+1)x^{2}+x = x(x+1), two distinct linear pieces, which is exactly the case partial fractions handles.

Write 1x(x+1)=Ax+Bx+1\frac{1}{x(x+1)} = \frac{A}{x} + \frac{B}{x+1} and clear denominators to get 1=A(x+1)+Bx1 = A(x+1) + Bx. Setting x=0x = 0 gives A=1A = 1, and setting x=−1x = -1 gives B=−1B = -1.

∫dxx2+x=∫dxx−∫dxx+1=ln⁡∣x∣−ln⁡∣x+1∣+C\int \frac{dx}{x^{2}+x} = \int \frac{dx}{x} - \int \frac{dx}{x+1} = \ln\left|x\right| - \ln\left|x+1\right| + C

Why the key shows a single logarithm

Two separate logs is a finished answer, but almost every printed solution compresses them. The quotient law ln⁡a−ln⁡b=ln⁡ab\ln a - \ln b = \ln\frac{a}{b} does the compressing.

ln⁡∣x∣−ln⁡∣x+1∣=ln⁡∣xx+1∣\ln\left|x\right| - \ln\left|x+1\right| = \ln\left|\frac{x}{x+1}\right|

If your work stops at the split form and the key shows the combined form, you have not made an error. Differentiating either one returns 1x2+x\frac{1}{x^{2}+x}.

The absolute value bars are doing real work

The integrand blows up at x = 0 and at x = -1, so it lives on three separate intervals. The bars let one formula cover all three, and strictly the constant of integration can differ from interval to interval. Drop the bars and the middle interval, negative 1 to 0, disappears, because the quotient x over x plus 1 is negative there.

The mistakes students make

All three of these produce a specific wrong answer, so they are easy to catch by differentiating what you wrote.

  • Splitting the fraction as 1x2+1x\frac{1}{x^{2}} + \frac{1}{x}, which is not what 1x2+x\frac{1}{x^{2}+x} equals. That path gives −1x+ln⁡∣x∣-\frac{1}{x} + \ln\left|x\right|, an antiderivative of a completely different function.
  • Mis-signing the second coefficient and getting B=1B = 1. The answer then comes out as ln⁡∣x(x+1)∣\ln\left|x(x+1)\right|, a product instead of a quotient, and its derivative is 2x+1x2+x\frac{2x+1}{x^{2}+x}.
  • Writing ln⁡xx+1\ln\frac{x}{x+1} with no bars, which is undefined on −1<x<0-1 < x < 0 even though the original integrand is perfectly well behaved there.

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 1/(x^2+x)?

Partial fractions turns it into ∫dxx−∫dxx+1\int\frac{dx}{x} - \int\frac{dx}{x+1}, and the quotient law folds the result into ln⁡∣xx+1∣+C\ln\left|\frac{x}{x+1}\right| + C.

Do I have to combine the two logs into one?

No. ln⁡∣x∣−ln⁡∣x+1∣+C\ln\left|x\right| - \ln\left|x+1\right| + C is the same function and earns the same credit. The combined form is a convention, not a requirement.

How do I know to use partial fractions here?

The integrand is a rational function, the numerator has lower degree than the denominator, and the denominator factors into distinct linear factors. That is the only partial fractions case the BC exam tests.