Choosing Antiderivatives for General Quadratic Radicands

Introduction

The integral of a reciprocal square root depends not only on the quadratic’s coefficients, but also on the shape of the quadratic. A completed square may produce a positive sum, a difference of squares, or a bounded difference. Each shape has its own antiderivative and real domain.

For a quadratic \(q(x)=Ax^2+Bx+C\) with \(A\ne 0\), the discriminant \(\Delta=B^2-4AC\) and the completed-square form provide a systematic way to make that choice. The first task is to determine where \(q(x)>0\), since the integrand \(1/\sqrt{q(x)}\) is real and defined only there.

Key idea: Complete the square, determine where the radicand is strictly positive, and then match its form to a logarithmic or inverse-sine formula. A zero of the radicand is excluded from the domain of the integrand.

Complete the Square and Identify the Domain

For \(A\ne 0\), define the centre \(h=-B/(2A)\). Completing the square gives

Ax^2+Bx+C = A(x-h)^2-\frac{\Delta}{4A}.

The sign of \(A\) determines whether the parabola opens upward or downward, while \(\Delta\) determines whether it has two, one, or no real roots. Together, these facts determine where the radicand is positive.

Coefficient and discriminant	Positive-radicand domain	Completed-square shape
\(A>0,\ \Delta<0\)	All real \(x\)	Positive sum: \(A((x-h)^2+a^2)\)
\(A>0,\ \Delta=0\)	\(x\ne h\)	Square: \(A(x-h)^2\)
\(A>0,\ \Delta>0\)	\(x<h-a\) or \(x>h+a\)	Difference: \(A((x-h)^2-a^2)\)
\(A<0,\ \Delta>0\)	\(h-a<x<h+a\)	Bounded difference: \((-A)(a^2-(x-h)^2)\)
\(A<0,\ \Delta\le 0\)	No real \(x\) with \(q(x)>0\)	The radicand is never positive

In the rows involving \(a\), take \(a=\sqrt{|\Delta|}/(2|A|)>0\). This choice makes the squared length in the completed-square form equal to \(a^2\). The strict inequalities in the domain column matter: at a root, the denominator is zero.

Choosing the Antiderivative

On any interval where \(q(x)>0\), an antiderivative is a function whose derivative equals \(1/\sqrt{q(x)}\). The formulas below are applied separately on each connected interval of the domain. The constant of integration may be different on different intervals.

Formula guide: For \(A>0\) and \(\Delta<0\), use an inverse-hyperbolic-sine or logarithmic form. For \(A>0\) and \(\Delta>0\), use a logarithmic form on either outside interval. For \(A<0\) and \(\Delta>0\), use an inverse-sine form on the bounded interval.

When \(A>0\) and \(\Delta<0\), the completed square is \(q(x)=A((x-h)^2+a^2)\), with \(a=\sqrt{-\Delta}/(2A)\). An antiderivative is

\int\frac{dx}{\sqrt{q(x)}} = \frac{1}{\sqrt A}\operatorname{arsinh}\!\left(\frac{x-h}{a}\right)+K = \frac{1}{\sqrt A} \ln\left|x-h+\sqrt{(x-h)^2+a^2}\right|+K.

The two displayed forms differ only by a constant, since \(\operatorname{arsinh}(u/a)=\ln((u+\sqrt{u^2+a^2})/a)\). The logarithm is real for every real \(x\) in this case.

When \(A>0\) and \(\Delta>0\), \(q(x)=A((x-h)^2-a^2)\), where \(a=\sqrt{\Delta}/(2A)\). The real domain consists of the two outside intervals, and on either one,

\int\frac{dx}{\sqrt{q(x)}} = \frac{1}{\sqrt A} \ln\left|x-h+\sqrt{(x-h)^2-a^2}\right|+K.

When \(A<0\) and \(\Delta>0\), the completed square is \(q(x)=(-A)(a^2-(x-h)^2)\), where \(a=\sqrt{\Delta}/(2|A|)\). Its positive domain is \(h-a<x<h+a\), and there

\int\frac{dx}{\sqrt{q(x)}} = \frac{1}{\sqrt{-A}} \arcsin\!\left(\frac{x-h}{a}\right)+K.

Here \(\arcsin\) is the principal inverse sine, with values in \([-\pi/2,\pi/2]\). Its argument lies strictly between \(-1\) and \(1\) on the integrand’s domain.

Why the Forms Work

The formulas follow by differentiating the corresponding expressions in the completed-square variable \(u=x-h\). For \(u^2+a^2\), differentiating the logarithmic expression gives the reciprocal square root:

\frac{d}{du}\ln\left|u+\sqrt{u^2+a^2}\right| = \frac{1+u/\sqrt{u^2+a^2}}{u+\sqrt{u^2+a^2}} = \frac{1}{\sqrt{u^2+a^2}}.

For \(u^2-a^2\), the same calculation applies wherever \(|u|>a\). The expression inside the absolute value is nonzero on both outside intervals, so the logarithm is defined there. For the bounded form, the chain rule gives the inverse-sine derivative. Including the constants from the coefficient \(A\) then produces the stated formulas.

Complete the square.
Write \(q(x)=A(x-h)^2-\Delta/(4A)\), with \(h=-B/(2A)\).

Find the real domain.
Use the signs of \(A\) and \(\Delta\) to determine where \(q(x)>0\); exclude every point where \(q(x)=0\).

Match the shape.
A positive sum gives a logarithmic or \(\operatorname{arsinh}\) form; a difference of squares gives a logarithm; a bounded difference gives an \(\arcsin\) form.

Check by differentiation.
Differentiate the proposed expression on the interval in question and simplify it to \(1/\sqrt{q(x)}\).

Worked Examples

Worked Example: Positive Sum with No Real Roots

Evaluate \(\int dx/\sqrt{2x^2+4x+5}\) and state its real domain.

Here \(A=2\), \(B=4\), and \(C=5\), so

\Delta=B^2-4AC=16-4(2)(5)=-24, \qquad h=-\frac{B}{2A}=-1.

Since \(A>0\) and \(\Delta<0\), the radicand is positive for all real \(x\). Completing the square gives

2x^2+4x+5 = 2(x+1)^2+3 = 2\left((x+1)^2+\frac32\right).

Thus \(a=\sqrt{3/2}\), and the logarithmic form gives

\int\frac{dx}{\sqrt{2x^2+4x+5}} = \frac{1}{\sqrt2} \ln\left|x+1+\sqrt{(x+1)^2+\frac32}\right|+K.

To check, put \(u=x+1\). The derivative of the logarithm is \(1/\sqrt{u^2+3/2}\), so the derivative of the whole expression is

\frac{1}{\sqrt2\sqrt{(x+1)^2+3/2}} = \frac{1}{\sqrt{2(x+1)^2+3}} = \frac{1}{\sqrt{2x^2+4x+5}}.

Worked Example: Bounded Radicand

Evaluate \(\int dx/\sqrt{9-4x^2}\) and state its real domain.

The coefficients are \(A=-4\), \(B=0\), and \(C=9\). Hence

\Delta=0^2-4(-4)(9)=144, \qquad h=0, \qquad a=\frac{\sqrt{144}}{2|{-4}|}=\frac32.

Because \(A<0\) and \(\Delta>0\), the radicand is positive exactly when \(-3/2<x<3/2\). Indeed, \(9-4x^2=4((3/2)^2-x^2)\). The inverse-sine formula gives

\int\frac{dx}{\sqrt{9-4x^2}} = \frac12\arcsin\!\left(\frac{2x}{3}\right)+K, \qquad -\frac32<x<\frac32.

Differentiating,

\frac12\cdot\frac{2/3}{\sqrt{1-4x^2/9}} = \frac{1}{3}\cdot\frac{3}{\sqrt{9-4x^2}} = \frac{1}{\sqrt{9-4x^2}}.

The derivative agrees with the integrand throughout the open interval; the endpoints are excluded because the radicand vanishes there.

Worked Example: Two Outside Intervals

Evaluate \(\int dx/\sqrt{x^2-4x+3}\), identifying both intervals on which the formula applies.

Here \(A=1\), \(B=-4\), \(C=3\), and

\Delta=(-4)^2-4(1)(3)=4, \qquad h=-\frac{-4}{2}=2, \qquad a=\frac{\sqrt4}{2}=1.

The completed square is \(x^2-4x+3=(x-2)^2-1\), which is positive when \(x<1\) or \(x>3\). Therefore, on either interval,

\int\frac{dx}{\sqrt{x^2-4x+3}} = \ln\left|x-2+\sqrt{(x-2)^2-1}\right|+K.

To verify it, let \(u=x-2\). The derivative of the logarithm is

\frac{1+u/\sqrt{u^2-1}}{u+\sqrt{u^2-1}} = \frac{1}{\sqrt{u^2-1}} = \frac{1}{\sqrt{x^2-4x+3}}.

This calculation holds on both intervals \(u<-1\) and \(u>1\). The absolute value makes the logarithm real on both, but it does not join the intervals into one domain: the integrand is undefined between the roots.

A Double Root Requires Special Care

If \(A>0\) and \(\Delta=0\), then \(q(x)=A(x-h)^2\). Its square root is \(\sqrt A\,|x-h|\), not \(\sqrt A(x-h)\) on the entire real line. The domain consists of the two intervals \(x<h\) and \(x>h\). On each interval, the antiderivative can be written as

\int\frac{dx}{\sqrt A\,|x-h|} = \begin{cases} -\dfrac{1}{\sqrt A}\ln|x-h|+K_-, & x<h,\\[6pt] \dfrac{1}{\sqrt A}\ln|x-h|+K_+, & x>h. \end{cases}

The constants are independent because the domain is disconnected. For example, if \(q(x)=4(x-1)^2\), the integrand is \(1/(2|x-1|)\). On \(x<1\), the correct derivative is that of \(-\tfrac12\ln|x-1|\), namely \(1/(2(1-x))\), which equals \(1/(2|x-1|)\).

Common pitfall: Replacing \(\sqrt{(x-h)^2}\) by \(x-h\) ignores the absolute value. For \(4(x-1)^2\), the incorrect expression \(\tfrac12\ln|x-1|\) has derivative \(-\tfrac12\) at \(x=0\), while the integrand there is \(1/2\). The negative sign in the antiderivative on \(x<1\) corrects the error.

Degenerate Cases

The classification above assumes \(A\ne0\), so that the radicand is genuinely quadratic. If \(A=0\), the expression is linear or constant instead. For \(B\ne0\), on the interval where \(Bx+C>0\),

\int\frac{dx}{\sqrt{Bx+C}}=\frac{2\sqrt{Bx+C}}{B}+K.

If \(A=B=0\) and \(C>0\), the integrand is the constant \(1/\sqrt C\), with antiderivative \(x/\sqrt C+K\). If the constant radicand is nonpositive, there is no real domain for the integrand.

Summary

For a genuine quadratic, completing the square and checking the signs of \(A\) and \(\Delta\) determines both the antiderivative type and its domain. A positive sum leads to a logarithmic or \(\operatorname{arsinh}\) expression, a difference of squares with \(A>0\) leads to a logarithm on two outside intervals, and a bounded difference with \(A<0\) leads to an inverse-sine expression between its roots. A double root must be handled separately on either side of the excluded point.

Check Your Understanding

For each question, identify the real domain before selecting or checking an antiderivative.

For \(q(x)=x^2+2x+5\), what are \(h\), \(\Delta\), and the real domain of \(1/\sqrt{q(x)}\)? Which form applies?
For \(q(x)=4-x^2\), determine the interval on which the integrand is real and choose the appropriate inverse-trigonometric form.
For \(q(x)=x^2-6x+8\), find the two domain intervals and state the logarithmic antiderivative form.
Why must the antiderivative for \(1/\sqrt{4(x-1)^2}\) be written separately on \(x<1\) and \(x>1\)?
Differentiate \(\frac{1}{\sqrt2}\operatorname{arsinh}(x/\sqrt2)\) and identify the corresponding radicand.