Split a rational function into integrable pieces

Try

Combine 1/x−1/(x+1) over a common denominator. The numerator becomes (x+1)−x=1, giving 1/[x(x+1)]. Reversing this calculation turns an awkward rational integrand into two logarithmic derivatives. State the excluded inputs before and after the transformation.

Predict

Predict whether the same two first-power terms suffice when the denominator contains (x+1)². A repeated factor allows more singular behavior. Omitting its higher-power term restricts the candidate identity before the coefficients are even calculated.

Experiment

Set 1/[x(x+1)]=A/x+B/(x+1). Multiplication by the denominator gives a polynomial identity.

Equation or substitution	Consequence
1=A(x+1)+Bx	identity for coefficients
x=0 in the polynomial identity	A=1
x=−1 in the polynomial identity	B=−1
Recombine the fractions	original numerator 1

Using the excluded input values in the multiplied polynomial identity is legitimate: the identity of polynomials extends to all real inputs. It does not assign values to the original rational expression at its poles. Explain that distinction before integrating.

Understand

Partial fractions decomposes a proper rational function after the denominator is factored. If the numerator’s degree is at least the denominator’s degree, perform polynomial division first. For (x²+1)/(x−1), division gives x+1+2/(x−1), whose primitive is x²/2+x+2ln|x−1|+C on each interval excluding 1.

A distinct linear factor contributes a constant numerator over that factor. A repeated factor (x−a)^m requires terms for every power from 1 through m. An irreducible real quadratic contributes a general linear numerator Ax+B; repeated quadratic factors similarly require all relevant powers. These forms allow a coefficient identity broad enough to represent the original expression. Special coefficients may later turn out to be zero, but they should not be assumed zero prematurely.

For 1/[x(x+1)²], write A/x+B/(x+1)+C/(x+1)². Multiplying out gives 1=A(x+1)²+Bx(x+1)+Cx. Substituting x=0 and x=−1 yields A=1 and C=−1; comparing the x² coefficient gives B=−1. The primitive is ln|x|−ln|x+1|+1/(x+1)+C₀. The last sign is positive because the derivative of 1/(x+1) is negative. Here C₀ is an integration constant, distinct from the decomposition coefficient C.

For a definite integral, first ensure the interval avoids every pole. On [1,2], ∫1/[x(x+1)]dx evaluates to ln(4/3). The result is positive, as expected from the positive integrand. If the interval crossed zero or −1, a single endpoint subtraction would not be justified; singularities require improper limits.

An irreducible quadratic often leads to both logarithmic and arctangent pieces after completing a square or separating the derivative of the denominator. For example, (2x+1)/(x²+1) integrates to ln(x²+1)+arctan x+C. Differentiate to verify both numerator contributions. More complicated repeated quadratics may require additional techniques beyond this introductory decomposition practice.

Apply

Check the degree, factor the denominator, write a complete candidate decomposition, solve coefficients and recombine before integrating. In written work, retain absolute values on logarithms unless the interval guarantees a positive argument. Explain why a partial-fraction identity does not remove the original singularities.

Continue

Next use trigonometric identities to expose a substitution. As with rational functions, algebraic restructuring often provides the essential step before integration.