A workshop in recognizing quadratic forms
Completing the square is useful because it reveals which standard antiderivative fits an integral. A quadratic denominator may become a shifted square plus a positive constant, leading to a logarithm and an arctangent. A quadratic radicand may become a positive-sign form or a bounded difference form, each with its own natural antiderivative.
The key is to show the square explicitly, then keep track of any scale factors introduced by the substitution. In this workshop, each example is checked by differentiating its antiderivative or by confirming the definite evaluation against its endpoints.
Linear numerators over irreducible quadratics
Suppose the denominator is a quadratic \(q(x)\), and the numerator is linear. The derivative \(q'(x)\) is also linear, so the numerator can be written as a constant multiple of \(q'(x)\), plus a constant. This separates the integral into a logarithmic part and a reciprocal-quadratic part.
For \(q(x)=ax^2+bx+c\), with \(a\ne 0\), choose \(\lambda=m/(2a)\) and \(\mu=n-mb/(2a)\). If \(q(x)>0\) on the interval of integration, the first part integrates to \(\lambda\ln q(x)\). If completing the square gives \(q(x)=a((x-h)^2+r^2)\), where \(a>0\) and \(r>0\), the remaining reciprocal has an arctangent antiderivative. The positive \(r\) is important: it fixes the scale in that formula.
Worked Example: A linear numerator over an irreducible quadratic
Evaluate \(\int \frac{3x+4}{x^2+2x+5}\,dx\). First complete the square:
The denominator’s derivative is \(2x+2\). Split the numerator:
Therefore the integral becomes
The first integral is \(\frac{3}{2}\ln(x^2+2x+5)\), since the denominator is positive. In the second, use the form \(\int \frac{du}{u^2+r^2}=\frac{1}{r}\arctan(u/r)+C\), with \(u=x+1\) and \(r=2\). Thus
To check, differentiate. The logarithmic term contributes \(\frac{3}{2}\frac{2x+2}{x^2+2x+5}\). The arctangent term contributes
Combining those contributions gives \(\frac{3x+3+1}{x^2+2x+5}=\frac{3x+4}{x^2+2x+5}\), the original integrand.
Why the reciprocal-quadratic formula has its scale factor
For \(r>0\), the basic arctangent form is
The factor \(1/r\) follows directly by differentiation. The chain rule contributes a factor \(1/r\) from the argument, while the denominator in the derivative of arctangent contributes the matching quadratic scale:
This check is a reliable way to avoid losing a scale factor when the completed square contains a constant other than \(1\).
Definite bounds on an irreducible denominator
A definite integral uses the same antiderivative as an indefinite one, but the Fundamental Theorem of Calculus requires the integrand to be continuous throughout the closed interval. A strictly positive completed-square denominator guarantees that condition. After finding an antiderivative \(F\), evaluate \(F(b)-F(a)\); do not substitute the bounds into the integrand instead.
Worked Example: An exact definite integral
Evaluate \(\int_{-2}^{0}\frac{dx}{x^2+4x+8}\). Complete the square:
This denominator is at least \(4\), so the integrand is continuous on \([-2,0]\). An antiderivative is
because differentiation gives
At the upper bound, \((0+2)/2=1\), and at the lower bound, \((-2+2)/2=0\). Therefore
The exact value is \(\pi/8\). The derivative check confirms the antiderivative, and the endpoint evaluations give the stated difference.
Radicals with a positive-sign quadratic
When a completed radicand has the form \(u^2+r^2\), with \(r>0\), its square root is real for every real \(u\). A standard antiderivative is
Here \(u+\sqrt{u^2+r^2}>0\) because \(r>0\), so the logarithm’s argument is positive. The absolute-value notation is harmless and often convenient when using related logarithmic formulas.
Worked Example: Integrating the square root of a positive-sign form
Evaluate \(\int\sqrt{x^2-2x+5}\,dx\). Complete the square and set \(u=x-1\):
Use the positive-sign formula with \(r=2\):
For a direct check, write \(u=x-1\) and \(s=\sqrt{u^2+4}\). Differentiating the displayed antiderivative gives \(\frac{1}{2}(s+u^2/s)+2/s=\frac{1}{2}(s+(u^2+4)/s)=s\). Since \(s=\sqrt{x^2-2x+5}\), the derivative is exactly the original integrand.
Bounded quadratic radicals and definite integrals
A difference form \(r^2-u^2\), with \(r>0\), is nonnegative only when \(-r\le u\le r\). On this interval, an antiderivative of its square root is
The inverse sine here uses its principal values in \([-\pi/2,\pi/2]\). In the interior \(-r<u<r\), differentiating the right side gives \(\frac{1}{2}\left(\sqrt{r^2-u^2}-\frac{u^2}{\sqrt{r^2-u^2}}+\frac{r^2}{\sqrt{r^2-u^2}}\right)=\sqrt{r^2-u^2}\). The antiderivative extends continuously to the endpoints, so it can be used for definite bounds there as well.
Worked Example: A definite integral of a bounded radical
Evaluate \(\int_{1}^{3}\sqrt{4-(x-1)^2}\,dx\). Set \(u=x-1\), so \(du=dx\). The bounds become \(u=0\) when \(x=1\), and \(u=2\) when \(x=3\). Thus
The radicand is nonnegative throughout these bounds. Apply the bounded-radical antiderivative with \(r=2\):
At \(u=2\), the square-root term is \(2\sqrt{0}=0\), and \(\arcsin(1)=\pi/2\), so \(G(2)=\frac{1}{2}(0+4\cdot\pi/2)=\pi\). At \(u=0\), both terms vanish, so \(G(0)=0\). Therefore
Differentiating \(G\) on \(0<u<2\) gives \(\sqrt{4-u^2}\), and the endpoint values give the exact definite integral.
A common scale-factor mistake
A frequent error is to complete the square correctly and then use the arctangent formula as if the constant inside the square were \(1\). For example, the following calculation is wrong:
The proposed right side differentiates to \(1/(1+(x+1)^2)\), which is not the integrand. The correct scale is \(r=2\), and the correct result is
Differentiating this expression gives \(1/((x+1)^2+4)\). In a definite integral, also check that the completed radicand or denominator stays in its required domain across the entire interval before applying the antiderivative and evaluating the bounds.
Choosing a route
A practical first step is to classify the completed square, rather than trying to integrate the original expression at sight.
- If a linear numerator accompanies a quadratic denominator, express the numerator using the denominator’s derivative and a constant.
- If the denominator becomes a shifted square plus a positive constant, use the arctangent form for the remaining reciprocal.
- If the radicand becomes \(u^2+r^2\), use the positive-sign square-root formula.
- If the radicand becomes \(r^2-u^2\), check the real domain \(|u|\le r\) and use the inverse-sine formula.
- For definite bounds, first check continuity on the full interval, then evaluate the antiderivative at the upper bound minus its value at the lower bound.
These checks connect the algebraic completion of the square to the conditions under which each antiderivative is valid. They also make the final derivative check straightforward.
Check Your Understanding
For each question, identify the completed form or condition before applying an antiderivative.
- Rewrite \(x^2+6x+13\) by completing the square. What reciprocal-quadratic form does it produce?
- Express \(5x+7\) as a constant multiple of the derivative of \(x^2+2x+5\), plus a constant.
- What condition on \(u\) makes \(\sqrt{9-u^2}\) real?
- Evaluate \(\int_{-1}^{1}\frac{dx}{(x+1)^2+1}\) using an antiderivative and state why the Fundamental Theorem of Calculus applies.
- Differentiate \(\frac{1}{3}\arctan(u/3)\). Which reciprocal-quadratic integrand does it produce?
