Integrating Shifted And Scaled Quadratic Forms

Quadratics in Vertex Form

Many rational integrals contain a denominator that is a positive scale times a square plus a positive constant: \(A(x-h)^2+k\) with \(A>0\) and \(k>0\). This vertex form has no real roots, because the expression is at least \(k>0\) for every real \(x\). The integral of its reciprocal is an inverse tangent. The method is one affine substitution that converts the denominator to \(k(u^2+1)\), and it reduces the problem to the basic integral \(\int\frac{du}{1+u^2}=\arctan u\).

Deriving the Formula

Let \(u=\sqrt{\frac Ak}\,(x-h)\). Then \(u^2=\frac Ak(x-h)^2\), so \(A(x-h)^2=ku^2\) and the denominator is \(ku^2+k=k(u^2+1)\). Also \(du=\sqrt{\frac Ak}\,dx\), so \(dx=\sqrt{\frac kA}\,du\). Therefore

The coefficient simplifies because \(\frac{\sqrt{k/A}}{k}=\frac{1}{\sqrt{A}\sqrt{k}}=\frac{1}{\sqrt{Ak}}\). Returning to \(x\) gives the stated formula.

Checking by Differentiation

By the chain rule, the derivative of \(\frac{1}{\sqrt{Ak}}\arctan u\) with \(u=\sqrt{A/k}\,(x-h)\) is \(\frac{1}{\sqrt{Ak}}\cdot\frac{1}{1+u^2}\cdot\sqrt{\frac Ak}=\frac{1}{k(1+u^2)}\), because \(\frac{1}{\sqrt{Ak}}\sqrt{\frac Ak}=\frac1k\). Since \(k(1+u^2)=k+A(x-h)^2\), this equals the integrand.

Worked Examples

A Reliable Procedure

Common Mistakes

The most frequent error is to put \(\sqrt{A/k}\) outside the arctangent or \(\frac{1}{\sqrt{Ak}}\) inside it. A second error is to use \(\frac1k\) as the outer factor, forgetting the chain-rule factor from \(du\). A third is to apply the formula when \(k\) is negative or zero, where the denominator has real roots and the integral is logarithmic.

Quick Reference

Interpreting the Result

The integrand is a bell-shaped curve centered at \(x=h\) with maximum height \(\frac1k\). The parameter \(A\) controls the width: larger \(A\) gives a narrower bump. The total area under the whole curve is \(\frac{\pi}{\sqrt{Ak}}\). For fixed \(k\), doubling \(A\) multiplies the area by \(\frac{1}{\sqrt2}\), reflecting a narrower curve of the same height. The antiderivative increases from \(-\frac{\pi}{2\sqrt{Ak}}\) to \(\frac{\pi}{2\sqrt{Ak}}\), with its steepest rise at the center.

A Numerical Spot Check

Check \(\int_0^1\frac{dx}{4(x-1)^2+1}\approx0.554\) with Simpson's rule on three points. The heights are \(\frac15=0.2\), \(\frac{1}{4\cdot0.25+1}=0.5\), and \(1\). Then \(\frac{0.5}{3}(0.2+4\cdot0.5+1)\approx0.5333\), within four percent of the exact value, as expected for a coarse rule on a curve with strong curvature.

Practice Problems with Answers

Try each problem before reading the answer. Every answer below was checked by differentiation or by numerical integration.

1. Find an antiderivative of \(\frac{1}{2(x-1)^2+3}\). Hint: \(A=2\), \(k=3\). Answer: \(\frac{1}{\sqrt6}\arctan\bigl(\sqrt{2/3}\,(x-1)\bigr)\)
2. Find an antiderivative of \(\frac{1}{4(x+2)^2+9}\). Hint: \(\sqrt{Ak}=6\). Answer: \(\frac16\arctan\frac{2(x+2)}{3}\)
3. Evaluate \(\int_0^1\frac{dx}{4(x-1)^2+1}\). Hint: the vertex is at \(x=1\). Answer: \(\frac12\arctan2\)
4. Evaluate \(\int_{-1}^{3}\frac{dx}{(x-1)^2+4}\). Hint: the interval is symmetric about the vertex. Answer: \(\frac\pi4\)

Where Vertex Form Comes From

In practice the denominator often arrives as \(ax^2+bx+c\) rather than \(A(x-h)^2+k\). Completing the square converts one to the other. For example, \(2x^2-8x+11=2(x^2-4x)+11=2(x-2)^2-8+11=2(x-2)^2+3\). Now \(A=2\), \(h=2\), \(k=3\), and the formula gives \(\int\frac{dx}{2x^2-8x+11}=\frac{1}{\sqrt6}\arctan\bigl(\sqrt{2/3}\,(x-2)\bigr)+C\). The next module develops this idea in full, but the vertex-form formula here is the target of that process, so it is worth mastering first.

The vertex form also reveals the geometry: the denominator has its minimum value \(k\) at \(x=h\), and the integrand has its maximum \(\frac1k\) there.

A Cross-Check by the Unscaled Formula

If \(A=k=1\), the formula collapses to \(\int\frac{dx}{(x-h)^2+1}=\arctan(x-h)\), the basic shifted arctangent. If \(A=1\) and \(k=a^2\), it gives the familiar \(\frac1a\arctan\frac{x-h}{a}\). These special cases are consistent with the general formula: \(\sqrt{Ak}=a\) and \(\sqrt{A/k}=\frac1a\). Testing a general formula on special cases is a quick way to catch a misplaced radical.

Choosing How to Present the Constants

The general formula can be written in other equivalent ways. Because \(\frac{1}{\sqrt{Ak}}\arctan\bigl(\sqrt{A/k}\,(x-h)\bigr)\) equals \(\frac{1}{\sqrt{Ak}}\arctan\frac{\sqrt{A}(x-h)}{\sqrt{k}}\), the inner factor may be rationalized, as in \(\arctan\frac{\sqrt{6}(x-1)}{3}\) for \(\sqrt{2/3}(x-1)\). The two expressions are equal, since \(\sqrt{2/3}=\frac{\sqrt6}{3}\). Teachers usually accept either form, but rationalizing the denominators makes numerical checks easier. When an answer is checked by a calculator, enter the inner factor as a decimal and compare the arctangent values directly. A common slip is to square or invert the inner factor, and a decimal check exposes that immediately.

Why the Area Does Not Depend on h

Shifting the center from \(h\) to \(0\) is the substitution \(t=x-h\), which has \(dt=dx\) and leaves the limits \(\pm\infty\) unchanged. So the total area is the same for every \(h\). A finite interval does depend on \(h\), because the interval moves relative to the center. For example, \(\int_{0}^{2}\frac{dx}{(x-1)^2+1}=\frac\pi2\) is twice \(\int_0^1\frac{dx}{(x-1)^2+1}=\frac\pi4\), while \(\int_0^2\frac{dx}{(x-5)^2+1}=\arctan5-\arctan3\approx0.124\) is much smaller because the interval sits away from the center.

Connection to Partial Fractions

In partial fraction decompositions, an irreducible quadratic factor produces terms of the form \(\frac{Bx+C}{A(x-h)^2+k}\). The constant-numerator case studied here handles the part of such a term that is not a derivative of the denominator. The remaining part, a multiple of the derivative \(2A(x-h)\), integrates to a logarithm. Together they account for every irreducible quadratic term, which is why this tutorial sits at the foundation of the partial fractions module and why the factors \(\frac{1}{\sqrt{Ak}}\) and \(\sqrt{A/k}\) appear again and again in decompositions with more complicated denominators.

Where This Fits in the Course

This tutorial belongs to the module “Partial Fractions Foundations”. The tutorial just before it is “Arctangent Terms From Constant Quadratic Numerators,” and the one that follows is “Templates For Repeated Irreducible Quadratics.” It is the general scaled and shifted form of the arctangent integral, a standard building block for partial fractions with irreducible quadratic factors.