Why repeated quadratic terms need two steps
A partial-fraction decomposition can contain terms whose denominators are powers of an irreducible quadratic, such as \((x^2+4)^2\) or \((x^2+1)^3\). When the numerator is linear, part of it may match the derivative of the quadratic. That part integrates directly by substitution; the rest requires an antiderivative for a power of the quadratic.
This tutorial develops that approach for definite integrals. A real irreducible quadratic has no real roots, so it does not vanish on the real interval of integration. Consequently, each individual term considered here is continuous on every finite interval, and its definite integral exists.
Split the numerator using the derivative
The derivative of \(Q(x)=ax^2+bx+c\) is \(Q'(x)=2ax+b\). Given the numerator \(Ax+B\), choose constants \(\lambda\) and \(\mu\) so that \(Ax+B=\lambda Q'(x)+\mu\). Matching the coefficient of \(x\) gives \(\lambda=\frac{A}{2a}\), and then matching constants gives \(\mu=B-\frac{Ab}{2a}\). Thus
For \(k\ge2\), the derivative-shaped part has an immediate antiderivative: since \(\frac{d}{dx}Q(x)^{1-k}=(1-k)Q'(x)Q(x)^{-k}\), its integral is a multiple of \(Q(x)^{1-k}\). The remaining constant-numerator term requires integrating \(Q(x)^{-k}\).
For a definite integral from \(u\) to \(v\), evaluate both contributions at the endpoints. In the formula above, the first contribution becomes \(\frac{A}{2a(1-k)}\bigl(Q(v)^{1-k}-Q(u)^{1-k}\bigr)\). The remaining task is to evaluate the definite integral of \(Q^{-k}\).
Reduce the quadratic to a standard form
Because \(Q\) is irreducible over the reals, its discriminant \(b^2-4ac\) is negative. Completing the square gives
Set \(z=x+\frac{b}{2a}\) and \(d=\frac{\sqrt{4ac-b^2}}{2a}\). Both \(a\) and \(d\) are positive, \(Q(x)=a(z^2+d^2)\), and \(dz=dx\). Therefore
For definite integrals, the transformed limits are \(u+\frac{b}{2a}\) and \(v+\frac{b}{2a}\). Define \(J_k(z)=\int\frac{dz}{(z^2+d^2)^k}\), with \(d>0\). The base case is \(J_1(z)=\frac{1}{d}\arctan(\frac{z}{d})\). Repeated powers can be obtained from a recurrence.
Why the recurrence works
To derive it, differentiate \(z(z^2+d^2)^{-(k-1)}\) using the product rule. The result is
Write the first term on the right over the denominator \((z^2+d^2)^k\), and use \(z^2=(z^2+d^2)-d^2\) in the second term. After collecting terms, the identity becomes
Integrating both sides gives the stated recurrence. This derivation also explains why the recurrence lowers the power by one: the derivative term supplies a boundary expression, and the remaining integral has denominator power \(k-1\). Starting with \(J_1\), it produces antiderivatives for every positive integer \(k\).
Worked Example: A symmetric interval with a squared quadratic
Worked Example: A symmetric interval with a squared quadratic
Evaluate \(\int_{-1}^{1}\frac{3x+5}{(x^2+4)^2}\,dx\).
Here \(Q(x)=x^2+4\), \(Q'(x)=2x\), and \(k=2\). Split the numerator as \(3x+5=\frac{3}{2}Q'(x)+5\). The integral is therefore
The first term has antiderivative \(-\frac{3}{2Q(x)}\). Since \(Q(-1)=Q(1)=5\), its definite contribution is \(-\frac{3}{10}-(-\frac{3}{10})=0\).
For the second term, use the \(k=2\) recurrence, or differentiate the resulting expression, to obtain \(\int\frac{dx}{(x^2+4)^2}=\frac{x}{8(x^2+4)}+\frac{1}{16}\arctan(\frac{x}{2})\). At \(x=1\) this is \(\frac{1}{40}+\frac{1}{16}\arctan(\frac12)\); at \(x=-1\) it is \(-\frac{1}{40}-\frac{1}{16}\arctan(\frac12)\). The difference is \(\frac{1}{20}+\frac18\arctan(\frac12)\). Multiplying by 5 gives
Check: differentiating \(\frac{x}{8(x^2+4)}+\frac{1}{16}\arctan(\frac{x}{2})\) gives \(\frac{1}{(x^2+4)^2}\); differentiating \(-\frac{3}{2(x^2+4)}\) gives \(\frac{3x}{(x^2+4)^2}\). Their weighted sum is exactly the original integrand. The endpoint subtraction above gives the stated definite value.
Worked Example: A cubic power after a shift
Worked Example: A cubic power after a shift
Evaluate \(\int_0^2\frac{2x+3}{((x-1)^2+1)^3}\,dx\).
Write \(Q(x)=(x-1)^2+1\), so \(Q'(x)=2(x-1)\) and \(2x+3=Q'(x)+5\). The integral separates into \(\int_0^2\frac{Q'(x)}{Q(x)^3}\,dx+5\int_0^2\frac{dx}{Q(x)^3}\). The first term has antiderivative \(-\frac{1}{2Q(x)^2}\). Since \(Q(0)=Q(2)=2\), its definite value is zero.
For the remaining integral, set \(z=x-1\), so the limits become \(-1\) and \(1\), and \(Q=z^2+1\). The recurrence gives
At \(z=1\), the three terms are \(\frac{1}{16}\), \(\frac{3}{16}\), and \(\frac{3\pi}{32}\), so \(J_3(1)=\frac14+\frac{3\pi}{32}\). At \(z=-1\), each term changes sign, so \(J_3(-1)=-\frac14-\frac{3\pi}{32}\). Hence the remaining integral is \(5\bigl(J_3(1)-J_3(-1)\bigr)\), and the answer is
Check: differentiating \(-\frac{1}{2Q(x)^2}+5J_3(x-1)\) gives \(Q'(x)/Q(x)^3+5/Q(x)^3=(2x+3)/Q(x)^3\). Its endpoint difference is the value just calculated.
Worked Example: A quadratic with a non-unit leading coefficient
Worked Example: A quadratic with a non-unit leading coefficient
Evaluate \(\int_0^1\frac{6x-1}{(2x^2+2x+3)^2}\,dx\).
Let \(Q(x)=2x^2+2x+3\). Then \(Q'(x)=4x+2\), and \(6x-1=\frac32Q'(x)-4\). Thus the derivative-shaped contribution is \(\frac32\int_0^1 Q'(x)Q(x)^{-2}\,dx\), which equals \(\left[-\frac{3}{2Q(x)}\right]_0^1\). Since \(Q(0)=3\) and \(Q(1)=7\), this contribution is \(-\frac{3}{14}+\frac12=\frac27\).
Complete the square: \(Q(x)=2((x+\frac12)^2+\frac54)\). With \(z=x+\frac12\) and \(d=\frac{\sqrt5}{2}\), the remaining quadratic-power integral is \(\int_0^1Q(x)^{-2}\,dx=\frac14[J_2(\frac32)-J_2(\frac12)]\). Here \(J_2(z)=\frac{z}{2d^2(z^2+d^2)}+\frac{1}{2d^3}\arctan(\frac{z}{d})\).
The rational part of \(J_2(\frac32)-J_2(\frac12)\) is \(\frac{6}{35}-\frac{2}{15}=\frac{4}{105}\). Its arctangent part is \(\frac{4}{5\sqrt5}\left(\arctan\frac{3}{\sqrt5}-\arctan\frac{1}{\sqrt5}\right)\). Including the factor \(\frac14\), we obtain
The original integral is \(\frac27\) minus four times this quantity. Therefore
Check: the antiderivative \(-\frac{3}{2Q(x)}-4\cdot\frac14J_2(x+\frac12)\) differentiates to \(\frac{3Q'(x)}{2Q(x)^2}-\frac{4}{Q(x)^2}=\frac{6x-1}{Q(x)^2}\). Evaluating it at \(1\) and \(0\) gives the displayed result.
A common mistake: matching the numerator incorrectly
The derivative of \(ax^2+bx+c\) is \(2ax+b\), not \(2ax\) unless \(b=0\). Ignoring the linear coefficient in the quadratic leads to an incorrect constant remainder and therefore an incorrect integral.
A second useful check is to differentiate any proposed antiderivative before evaluating its endpoints. This catches errors in powers and constants in the recurrence, as well as mistakes in the numerator split.
Using the method reliably
Write the term as \((Ax+B)/Q(x)^k\), with \(Q(x)=ax^2+bx+c\) irreducible and \(a>0\).
Use \(Ax+B=\frac{A}{2a}Q'(x)+(B-\frac{Ab}{2a})\).
For \(k\ge2\), use the antiderivative \(\frac{A}{2a(1-k)}Q(x)^{1-k}\), then take its endpoint difference.
Transform \(Q\) to \(a(z^2+d^2)\), adjust the limits, and use the recurrence for the required power.
Differentiate the constructed antiderivative and confirm that it reproduces the original integrand before evaluating endpoint values.
This method is useful because it treats the two sources of difficulty separately. The numerator’s derivative-shaped part is handled by substitution, while the constant remainder is handled by the standard quadratic-power recurrence. For a complete partial-fraction decomposition, apply the procedure to each repeated irreducible-quadratic term and add the resulting definite integrals.
Check Your Understanding
Use the numerator split and quadratic-power recurrence to answer these questions.
- For \(Q(x)=3x^2-6x+8\), find \(Q'(x)\) and complete the square.
- Write \(5x+4\) as a multiple of the derivative of \(Q(x)=x^2+2x+5\) plus a constant.
- What is an antiderivative of \(Q'(x)/Q(x)^4\) when \(Q(x)\) is positive?
- In the recurrence for \(J_k\), which lower-power integral appears on the right-hand side?
- Why must the transformed endpoints change when setting \(z=x+\frac{b}{2a}\) in a definite integral?
