Why the Denominator Gives You a Clue
A rational function is a quotient of polynomials. When its denominator factors over the real numbers, those factors provide a useful guide to the form of a partial-fraction decomposition. Linear factors lead to terms with constant numerators; irreducible quadratic factors lead to terms with linear numerators. Repeated factors require a term for each power up to the factor’s multiplicity.
Recognizing those forms helps you decide whether partial fractions are a natural strategy for an integral. It also suggests the antiderivatives to expect: logarithms for linear factors, and logarithms and inverse tangents for irreducible quadratics.
If the numerator’s degree is not less than the denominator’s, these forms cannot be applied directly to the whole rational function. For now, the focus is on reading factors of a denominator and predicting the decomposition they call for.
The Forms Signalled by Linear Factors
A factor \(x-r\) in the denominator contributes a term of the form \(A/(x-r)\), where \(A\) is a constant. If the factor is repeated, its successive powers must all be represented. For example, a denominator containing \((x-r)^3\) calls for terms with denominators \(x-r\), \((x-r)^2\), and \((x-r)^3\).
When the power in a term is one, integration produces a logarithm. Higher powers can be integrated using the power rule after writing the denominator as a negative power. A useful sign check is that \(\int (x-r)^{-2}\,dx=-(x-r)^{-1}+C\).
The Forms Signalled by Irreducible Quadratics
An irreducible quadratic over the real numbers is a quadratic polynomial that has no real roots. Write one as \(q(x)=ax^2+bx+c\), with \(a\ne 0\) and \(b^2-4ac<0\). Because its numerator may need both a constant part and a part proportional to \(x\), a factor \(q(x)\) contributes a term \((Ax+B)/q(x)\), with constants \(A\) and \(B\).
For a repeated factor \(q(x)^m\), include one such term for every power from one through \(m\).
The numerator is linear because it must have degree less than the quadratic denominator. This choice also supports integration: for a first-power term, its numerator can be separated into a multiple of \(q'(x)\) and a constant. The derivative part gives a logarithm, while the constant part leads to an inverse tangent.
Why the First-Power Forms Integrate as Expected
For a linear factor, the logarithm follows directly from the derivative of a logarithm: \(\frac{d}{dx}\ln|x-r|=1/(x-r)\), whenever \(x\ne r\).
For an irreducible quadratic \(q(x)=ax^2+bx+c\), let \(\Delta=4ac-b^2\), which is positive. Its derivative is \(q'(x)=2ax+b\). Any linear numerator \(Ax+B\) can be written as a multiple of \(q'(x)\) plus a constant:
The first part integrates to a logarithm. For the constant part, the following formula gives an inverse tangent:
To verify it, note that \(4a q(x)=(2ax+b)^2+\Delta\). Differentiating the proposed right-hand side gives \(4a/((2ax+b)^2+\Delta)=1/q(x)\). Thus the formula is valid wherever the quadratic is nonzero; an irreducible real quadratic has no real zeros.
Worked Examples
Worked Example: Distinct Linear Factors
Integrate \(\frac{5x+1}{(x-1)(x+2)}\). The denominator has two distinct linear factors, so try a constant numerator over each:
Multiply by \((x-1)(x+2)\) to get \(5x+1=A(x+2)+B(x-1)\). At \(x=1\), \(6=3A\), so \(A=2\). At \(x=-2\), \(-9=-3B\), so \(B=3\). Therefore:
Check by differentiating: the derivative is \(2/(x-1)+3/(x+2)\). Combining those fractions gives \([2(x+2)+3(x-1)]/[(x-1)(x+2)]=(5x+1)/[(x-1)(x+2)]\), as required.
Worked Example: A Repeated Linear Factor
Integrate \(\frac{1}{x(x+1)^2}\). The factor \(x+1\) is repeated, so the decomposition needs terms over both its first and second powers:
Multiplying through by \(x(x+1)^2\) gives \(1=A(x+1)^2+Bx(x+1)+Cx\). Setting \(x=0\) gives \(A=1\); setting \(x=-1\) gives \(C=-1\). Comparing the \(x^2\) coefficients gives \(A+B=0\), so \(B=-1\). Thus:
Differentiate to check: the derivative is \(1/x-1/(x+1)-1/(x+1)^2\). With common denominator \(x(x+1)^2\), its numerator is \((x+1)^2-x(x+1)-x=1\). This recovers the integrand.
Worked Example: An Irreducible Quadratic
Integrate \(\frac{3x+5}{x^2+2x+5}\). The denominator has discriminant \(2^2-4(1)(5)=-16<0\), so it is irreducible over the real numbers. Its derivative is \(2x+2\), and the numerator splits as \(3x+5=\frac32(2x+2)+2\). Therefore:
Here \(\Delta=4(1)(5)-2^2=16\). Applying the inverse-tangent formula to the remaining integral gives:
For the check, differentiate the logarithmic term to obtain \(\frac32(2x+2)/(x^2+2x+5)\). Differentiating the inverse-tangent term gives \(2/(x^2+2x+5)\). Their sum is \((3x+3+2)/(x^2+2x+5)\), the original integrand.
Worked Example: A Repeated Irreducible Quadratic
Integrate \(\frac{1}{(x^2+1)^2}\). The denominator is a repeated irreducible quadratic. Its partial-fraction template would allow a linear numerator over each power; in this case the integrand already consists of a constant numerator over the second power. A suitable antiderivative is:
To verify it, differentiate the first term to get \((1-x^2)/(2(x^2+1)^2)\), and the second to get \(1/(2(x^2+1))\). Writing the second derivative over the common denominator gives \((x^2+1)/(2(x^2+1)^2)\). Adding yields \(2/(2(x^2+1)^2)=1/(x^2+1)^2\).
A Common Mistake: Omitting Powers
A repeated factor does not call for only one partial-fraction term. For instance, the incorrect setup
omits the term \(C/(x+1)\). The correct form includes every power of the repeated factor:
As the second worked example showed, the missing term is not necessarily zero: its coefficient is \(B=-1\). Omitting it prevents the decomposition from matching the original rational function.
Using the Pattern to Choose a Strategy
Partial fractions are a natural candidate when a rational integrand is proper and its denominator factors into real linear factors and irreducible quadratics. The decomposition then turns one rational function into a sum of simpler terms with recognizable antiderivative forms. The factor pattern predicts the setup; the numerator and coefficient calculations determine the exact terms.
The method is not automatically the best choice for every rational function. Factoring may be difficult, and a numerator may already align with the derivative of the denominator. Even when partial fractions are suitable, repeated quadratic terms can require more involved integration than first-power terms. Recognizing the denominator’s structure helps you anticipate that work before committing to a method.
Check Your Understanding
Use the denominator structure to answer these questions.
- What partial-fraction terms should be included for a denominator containing \((x-2)^3\)?
- What numerator form belongs over an irreducible quadratic factor \(x^2+4\)?
- Why must a repeated factor contribute terms for each power up to its multiplicity?
- What two types of antiderivative can arise from a first-power irreducible quadratic term?
- Why does the form in this tutorial apply directly only when the rational function is proper?
