Why a Recurrence Helps
Repeated irreducible quadratic factors lead to integrals with increasingly high powers in the denominator. When the numerator is a constant, completing each integral from scratch can be inefficient. A recurrence gives a systematic way to reduce the power: an integral with exponent \(n\) is expressed using one with exponent \(n-1\), together with an explicit algebraic term.
This method complements the earlier material on derivative-matched numerators over quadratic powers. There, a numerator proportional to the derivative of the quadratic leads directly to a logarithmic antiderivative. Here the numerator is constant, so the reduction ultimately reaches an arctangent.
The condition \(a>0\) ensures that \(x^2+a^2\) is never zero for real \(x\), and it makes the base antiderivative take the standard arctangent form. The recurrence below applies for every integer \(n\geq 2\); the starting case is \(n=1\).
The Base Case and the Reduction Formula
For \(n=1\), the standard arctangent antiderivative gives the starting point. For higher powers, the recurrence reduces the exponent by one.
To use the formula, calculate the algebraic term for the current exponent, multiply the preceding antiderivative by its stated coefficient, and add the results. The appearance of \(a^2\) in both coefficients matters: the recurrence is not obtained simply by replacing \(x^2+1\) with \(x^2+a^2\) in a formula that has no scale parameter.
Why the Recurrence Works
The key is to differentiate \(x/(x^2+a^2)^{n-1}\). This choice produces both the power-\(n\) denominator we want and a term involving the power-\((n-1)\) denominator. Set \(D=x^2+a^2\) temporarily to keep the algebra readable.
Using the product rule and the chain rule, $$ \frac{d}{dx}\left(xD^{-(n-1)}\right) =D^{-(n-1)}-2(n-1)x^2D^{-n}. $$
Substitution gives $$ D^{-(n-1)}-2(n-1)(D-a^2)D^{-n} =(3-2n)D^{-(n-1)}+2(n-1)a^2D^{-n}. $$
Rearrange the last equality: $$ D^{-n}=\frac{1}{2(n-1)a^2}\frac{d}{dx}\left(xD^{-(n-1)}\right) +\frac{2n-3}{2(n-1)a^2}D^{-(n-1)}. $$
The derivative term integrates to \(xD^{-(n-1)}/[2(n-1)a^2]\), while the remaining term integrates to its coefficient times \(I_{n-1}\). This is the stated recurrence.
This derivation explains the structure of the result, not just its coefficients. Differentiating the algebraic term in the recurrence produces the new integrand, except for a leftover multiple of the preceding power. The recurrence supplies exactly that leftover through \(I_{n-1}\).
Worked Example: Finding the Power-Two Antiderivative
Worked Example: Finding the Power-Two Antiderivative
Find \(\int \frac{dx}{(x^2+1)^2}\). Here \(a=1\), \(n=2\), and \(n-1=1\). The algebraic term is $$ \frac{x}{2(1)^2(1)(x^2+1)^1}=\frac{x}{2(x^2+1)}. $$ The coefficient of \(I_1\) is $$ \frac{2(2)-3}{2(1)^2(1)}=\frac{1}{2}. $$ Since \(I_1=\arctan x\), the recurrence gives $$ I_2=\frac{x}{2(x^2+1)}+\frac{1}{2}\arctan x+C. $$ Check by differentiating. The derivative of the first term is \((1-x^2)/[2(x^2+1)^2]\), and the derivative of the second is \(1/[2(x^2+1)]\). With the common denominator \(2(x^2+1)^2\), their sum is $$ \frac{1-x^2}{2(x^2+1)^2}+\frac{x^2+1}{2(x^2+1)^2} =\frac{1}{(x^2+1)^2}. $$ Thus the derivative matches the original integrand.
Working Up to Higher Powers
Once \(I_2\) is known, the same recurrence produces \(I_3\), then \(I_4\), and so on. At each step, use the preceding result rather than restarting the derivation. The coefficient of \(I_{n-1}\) is positive for all \(n\geq2\), since \(2n-3\geq1\), and the algebraic term has denominator \((x^2+a^2)^{n-1}\).
Worked Example: Building the Power-Three Antiderivative
Find \(\int \frac{dx}{(x^2+1)^3}\), using the result for \(I_2\). For \(n=3\), the coefficient of the algebraic term is $$ \frac{1}{2(1)^2(3-1)}=\frac{1}{4}, $$ and the coefficient of \(I_2\) is $$ \frac{2(3)-3}{2(1)^2(3-1)}=\frac{3}{4}. $$ Therefore, $$ I_3=\frac{x}{4(x^2+1)^2}+\frac{3}{4}I_2 =\frac{x}{4(x^2+1)^2}+\frac{3x}{8(x^2+1)}+\frac{3}{8}\arctan x+C. $$ For a check, differentiate the first two terms: $$ \frac{d}{dx}\left(\frac{x}{4(x^2+1)^2}\right) =\frac{1-3x^2}{4(x^2+1)^3}, $$ $$ \frac{d}{dx}\left(\frac{3x}{8(x^2+1)}\right) =\frac{3(1-x^2)}{8(x^2+1)^2} =\frac{3(1-x^4)}{8(x^2+1)^3}. $$ The arctangent term has derivative \(3/[8(x^2+1)] = 3(x^2+1)^2/[8(x^2+1)^3]\). Adding these numerators over the common denominator \(8(x^2+1)^3\) gives $$ 2(1-3x^2)+3(1-x^4)+3(x^2+1)^2 =8. $$ Hence \(I_3'=8/[8(x^2+1)^3]=1/(x^2+1)^3\), as required.
A Scaled Quadratic and a Definite Integral
The parameter \(a\) changes the coefficients in the recurrence, and it also changes the base arctangent. A definite integral can be evaluated by using the resulting antiderivative at the two endpoints. The following example displays both effects.
Worked Example: Evaluating a Definite Integral with \(a=2\)
Evaluate \(\int_0^2 \frac{dx}{(x^2+4)^2}\). Set \(a=2\) and \(n=2\). The recurrence gives $$ I_2=\frac{x}{2(2)^2(x^2+4)}+\frac{1}{2(2)^2}I_1 =\frac{x}{8(x^2+4)}+\frac{1}{8}I_1. $$ The base case is \(I_1=\frac{1}{2}\arctan(x/2)\), so $$ I_2=\frac{x}{8(x^2+4)}+\frac{1}{16}\arctan\!\left(\frac{x}{2}\right). $$ At \(x=2\), the algebraic term is \(2/[8(8)]=1/32\), and the arctangent term is \(\arctan(1)/16=\pi/64\). At \(x=0\), both terms are zero. Therefore, $$ \int_0^2 \frac{dx}{(x^2+4)^2} =\left[\frac{x}{8(x^2+4)}+\frac{1}{16}\arctan\!\left(\frac{x}{2}\right)\right]_0^2 =\frac{1}{32}+\frac{\pi}{64} =\frac{2+\pi}{64}. $$ As a derivative check, the derivative of the algebraic term is \((4-x^2)/[8(x^2+4)^2]\), and the derivative of the arctangent term is \(1/[8(x^2+4)]\). Their sum is $$ \frac{4-x^2}{8(x^2+4)^2}+\frac{x^2+4}{8(x^2+4)^2} =\frac{1}{(x^2+4)^2}. $$ The antiderivative is therefore correct, and its endpoint difference gives the stated definite integral.
How This Fits Repeated Quadratic Terms
In integrating rational functions, partial fractions can produce terms whose denominators contain repeated irreducible quadratics. For a quadratic written in the form \(x^2+a^2\), the numerator over one of its powers is linear. The part proportional to \(x\) matches the derivative of the quadratic up to a constant, while the constant-numerator part leads to the integrals treated here.
For example, once a decomposition has produced a constant multiple of \(1/(x^2+a^2)^n\), the recurrence reduces its integral to the power-\((n-1)\) case. Repeating this process eventually reaches \(I_1\), which is an arctangent. Thus the method supplies an organized way to integrate every constant-numerator term in a repeated quadratic tower. The decomposition itself is not part of this recurrence; the recurrence begins after the term to be integrated has been identified.
A Common Coefficient Mistake
A frequent error is to use \(2n-1\) in place of \(2n-3\) in the coefficient multiplying \(I_{n-1}\). Testing the incorrect version at \(n=2\), \(a=1\), would give $$ I_2^{\mathrm{wrong}}=\frac{x}{2(x^2+1)}+\frac{3}{2}\arctan x. $$ Differentiating this expression gives $$ \frac{1-x^2}{2(x^2+1)^2}+\frac{3}{2(x^2+1)} =\frac{2+x^2}{(x^2+1)^2}, $$ which is not \(1/(x^2+1)^2\). The correction is to return to the derived factor \(2n-3\); when \(n=2\), that factor is \(1\), so the coefficient is \(1/2\), not \(3/2\).
Check Your Understanding
Use the recurrence and the base case to answer these questions. Check antiderivatives by differentiating where appropriate.
- State the recurrence for \(I_n\), including the conditions on \(a\) and \(n\).
- Use the recurrence to find an antiderivative of \(1/(x^2+1)^2\).
- For \(a=3\) and \(n=2\), what is the coefficient multiplying \(I_1\)?
- Why does differentiating \(x/(x^2+a^2)^{n-1}\) produce both the power-\(n\) and power-\((n-1)\) terms?
- What is the base antiderivative \(I_1\), and why must the recurrence be applied only for \(n\geq2\)?
