Why Reduce a Squared Quadratic Denominator?
Repeated powers of an irreducible quadratic can produce integrals that look increasingly complicated. For a squared denominator, the integral \(\int \frac{du}{(u^2+a^2)^2}\) does not have the same immediate inverse-tangent form as the reciprocal first power. Higher powers appear in the same way: direct expansion is not useful, but each integral can be related to one with a lower power.
The reduction formula makes that relationship explicit. Once the first-power integral is known, it gives the squared case and then each successive integer power. Completing the square lets us use the same result when the denominator is a general irreducible quadratic.
The Base Integral and the Recurrence
The base integral follows from the inverse-tangent derivative. For \(a>0\), \(\frac{d}{du}\arctan(u/a)=a/(u^2+a^2)\), so \(\int \frac{du}{u^2+a^2}=\frac{1}{a}\arctan(u/a)+C\). The reduction formula expresses the integral for \(n\ge 2\) using this base case and the integral with exponent \(n-1\).
Why the Reduction Formula Works
The key is to differentiate \(u/(u^2+a^2)^{n-1}\). Its derivative contains both a reciprocal \(n\)th power and a reciprocal \((n-1)\)th power. Rearranging that identity isolates the \(n\)th power.
Use the product rule, writing the denominator as a negative power.
Replace \(u^2\) by \((u^2+a^2)-a^2\), so both denominator powers appear.
Divide by \(2a^2(n-1)\), then integrate both sides with respect to \(u\).
For the full algebra, set \(D=u^2+a^2\). Differentiating gives
The last equality uses \(u^2=D-a^2\). Rearranging the identity yields
Integrating this equation proves the recurrence. The restriction \(n\ge 2\) is necessary: the derivation divides by \(n-1\), which is zero for \(n=1\). That case is supplied separately by the base integral.
The Squared-Denominator Case
Set \(n=2\) in the recurrence. The coefficients become \(1/(2a^2)\) and \(1/(2a^2)\), respectively. Substituting the base integral then gives a direct formula for the reciprocal square.
Worked Example: Integrating a Reciprocal Squared Quadratic
Evaluate \(\int \frac{du}{(u^2+9)^2}\). Here \(a=3\), so \(a^2=9\). With \(n=2\), the reduction formula gives
To check, differentiate the two terms. The derivative of the first is
and the derivative of the second is \(\frac{1}{54}\cdot\frac{1/3}{1+(u/3)^2}=\frac{1}{18(u^2+9)}\). Putting the second derivative over the common denominator gives \(\frac{u^2+9}{18(u^2+9)^2}\). Adding the derivatives yields \(\frac{9-u^2+u^2+9}{18(u^2+9)^2}=\frac{1}{(u^2+9)^2}\), as required.
For any \(a>0\), the corresponding formula is
Reducing a Higher Power
The recurrence can be applied repeatedly. At each step, the exponent decreases by one until the base integral is reached. For instance, setting \(n=3\) gives \(\int \frac{du}{(u^2+a^2)^3} =\frac{u}{4a^2(u^2+a^2)^2}+\frac{3}{4a^2}\int\frac{du}{(u^2+a^2)^2}\). Substituting the squared-denominator formula produces an explicit antiderivative.
Worked Example: Reducing a Reciprocal Cube
Evaluate \(\int \frac{du}{(u^2+1)^3}\). Take \(a=1\) and \(n=3\). The recurrence gives
The squared case with \(a=1\) is \(\int\frac{du}{(u^2+1)^2}=\frac{u}{2(u^2+1)}+\frac12\arctan(u)+C\). Therefore,
To check, differentiate the first term to obtain \(\frac{1-3u^2}{4(u^2+1)^3}\). Differentiating the second gives \(\frac{3(1-u^2)}{8(u^2+1)^2}\), and differentiating the third gives \(\frac{3}{8(u^2+1)}\). With common denominator \(8(u^2+1)^3\), the three numerators are \(2(1-3u^2)\), \(3(1-u^2)(u^2+1)=3(1-u^4)\), and \(3(u^2+1)^2=3(u^4+2u^2+1)\). Their sum is \(8\), so the derivative is \(1/(u^2+1)^3\).
Applying the Formula After Completing the Square
A general irreducible quadratic can be put into the required form when its leading coefficient is positive. Let \(q(x)=Ax^2+Bx+C\), where \(A>0\) and \(4AC-B^2>0\). Completing the square gives
Set \(u=x+\frac{B}{2A}\) and \(a=\frac{\sqrt{4AC-B^2}}{2A}\), which is positive under these assumptions. Since \(du=dx\), we have \(q(x)=A(u^2+a^2)\). For the squared denominator, this means \(\frac{1}{q(x)^2}dx=\frac{1}{A^2}\frac{du}{(u^2+a^2)^2}\). Both the shift in the variable and the factor \(A^{-2}\) matter.
Worked Example: Completing the Square in a Squared Denominator
Evaluate \(\int \frac{dx}{(2x^2+4x+6)^2}\). First complete the square:
Let \(u=x+1\). Then \(du=dx\), and the denominator squared is \(4(u^2+2)^2\). Thus
For a derivative check, write \(u=x+1\). The derivative of the first term is \(\frac{2-u^2}{16(u^2+2)^2}\). The derivative of the inverse-tangent term is \(\frac{1}{16\sqrt{2}}\cdot\frac{1/\sqrt{2}}{1+u^2/2} =\frac{1}{16(u^2+2)}\). Combining them gives \[ \frac{2-u^2}{16(u^2+2)^2} +\frac{u^2+2}{16(u^2+2)^2} =\frac{1}{4(u^2+2)^2} =\frac{1}{(2x^2+4x+6)^2}. \] The derivative matches the integrand.
Common Pitfall: Losing the Leading-Coefficient Factor
A common error is to complete the square correctly but then treat \(A(u^2+a^2)\) as though it were just \(u^2+a^2\). Squaring the denominator squares the factor \(A\), so the integral acquires a factor \(1/A^2\). In the preceding example, omitting that factor would give an antiderivative four times too large.
Worked Example: Diagnosing a Missing Scale Factor
For \(2x^2+4x+6=2((x+1)^2+2)\), an incorrect first step would be \(\frac{1}{(2x^2+4x+6)^2}=\frac{1}{((x+1)^2+2)^2}\). In fact,
The incorrect antiderivative would therefore differentiate to \(1/((x+1)^2+2)^2\), which is four times the intended integrand. The correction is to keep \(1/4\) through the substitution and the integration, as in the previous calculation; its derivative is \(1/[4((x+1)^2+2)^2]\), exactly the original integrand.
There are two other conditions to keep in view. The recurrence requires \(n\ge2\), while the base case handles \(n=1\). Also, this form assumes \(a>0\); if the completed-square constant is not positive, the denominator is not \(u^2+a^2\) with a positive real \(a\), and this particular formula does not apply as stated.
Using the Recurrence Efficiently
For a reciprocal power with integer exponent \(n\ge2\), the practical procedure is to identify \(a\), apply the recurrence once, and continue reducing until the exponent reaches one. At each stage, retain the coefficient multiplying the lower-power integral. The base case supplies the inverse tangent, and the recurrence supplies the rational terms.
This method matters because it turns a family of integrals into one reusable calculation. It also gives a useful check: differentiating the reduction identity must recover \(1/(u^2+a^2)^n\), and completing the square must preserve every constant multiplier. Keeping those two checks visible prevents the most common algebraic and scaling errors.
Check Your Understanding
Use the recurrence and the stated conditions to answer these questions. Check each proposed antiderivative by differentiation.
- What restrictions on \(a\) and \(n\) are needed for the reduction formula, and why is \(n=1\) treated separately?
- Write the recurrence for \(\int \frac{du}{(u^2+4)^3}\), including its coefficient multiplying the lower-power integral.
- Use the squared-denominator formula to evaluate \(\int \frac{du}{(u^2+16)^2}\).
- Complete the square in \(3x^2+6x+9\), and state the factor that must be included when converting its squared reciprocal to the standard form.
- Explain how differentiation checks whether a missing leading-coefficient factor has changed an answer.
