Combining Linear and Quadratic Factors in One Integral

Why Mixed Factors Require More Than One Antiderivative

A rational function may have a denominator that factors into both real linear factors and irreducible quadratic factors. Partial fractions separate the expression into simpler pieces, but those pieces do not all integrate in the same way: a linear denominator produces a logarithm, while an irreducible quadratic can produce both a logarithm and an inverse tangent.

The main task is therefore to keep the algebra and integration connected. Each partial fraction must have the correct numerator for its denominator, and the quadratic terms must be arranged so their numerators match the derivative of the quadratic or leave a constant remainder.

The Decomposition Pattern

For a proper rational function whose denominator is a product of distinct real linear factors and distinct irreducible quadratic factors, the partial-fraction form has a constant numerator over each linear factor and a linear numerator over each quadratic factor. “Proper” means the degree of the numerator is less than the degree of the denominator. If a rational function is not proper, polynomial division is needed first.

Here, the \(r_i\) are real numbers, each \(Q_j\) is irreducible over the reals, and the constants \(A_i,B_j,C_j\) are determined by algebra. The linear numerator \(B_jx+C_j\) is necessary: it is general enough to contain a multiple of \(Q_j'(x)\), as well as a constant remainder.

For a linear factor, the relevant antiderivative is \(\int \frac{A}{x-r}\,dx=A\ln|x-r|+C\), valid on any interval not containing \(r\). For a quadratic factor, the completed-square form leads to the logarithmic and inverse-tangent terms.

How a Quadratic Term Splits

Let \(Q(x)=ax^2+bx+c\) be irreducible over the reals, and let its numerator be \(Bx+C\). Since \(Q'(x)=2ax+b\), choose a multiple of \(Q'\) that has the same coefficient of \(x\) as the numerator. The leftover is a constant:

Completing the square gives \(Q(x)=a\big((x-h)^2+k^2\big)\), where \(h=-\frac{b}{2a}\) and \(k=\frac{\sqrt{4ac-b^2}}{2|a|}>0\). Thus the first part integrates as a logarithm, and the constant remainder integrates as an inverse tangent.

Why the Quadratic Formula Works

The decomposition of the numerator is an identity, not a guess. To verify the integration result, differentiate its two terms. The derivative of the logarithmic term is \(\frac{B}{2a}\frac{Q'}{Q}\). Also, because \(Q=a((x-h)^2+k^2)\), the chain rule gives

Multiplying this derivative by the constant remainder \(C-\frac{Bb}{2a}\), and adding the logarithmic derivative, gives

The numerator equality follows from \(Q'(x)=2ax+b\). Therefore differentiation recovers the original quadratic fraction, which proves the formula on any interval where the integrand is defined.

Worked Examples

A Common Scaling Error

A frequent mistake is to integrate \(\frac{1}{(x-1)^2+4}\) as \(\arctan(x-1)\). Differentiating that proposed answer gives \(\frac{1}{1+(x-1)^2}\), which is not the original integrand. The \(4\) changes the scale inside the inverse tangent and contributes a factor outside it.

because differentiating the right-hand side gives \(\frac{1}{(x-1)^2+4}\). A reliable check is to differentiate every inverse-tangent term before accepting the result.

Putting the Method Together

For a mixed rational integral, the workflow is systematic. The decomposition handles the algebraic structure of the denominator; the shape of each numerator then identifies the appropriate antiderivative.

Mixed-factor integrals are useful practice because they combine partial-fraction algebra with two standard antiderivative forms. The key is not to treat every term alike: linear factors lead to logarithms, while irreducible quadratics may require both a logarithm and an inverse tangent. Keep track of the intervals where the integrand is defined, and include absolute values in logarithms of real linear factors.