Why These Integrals Need More Than One Idea
Some integrals contain both a rational factor and the square root of a quadratic expression. A useful first step is to complete the square in the radicand. The resulting form can reveal a derivative that is already present in the numerator, or show that the integrand can be rewritten using a standard antiderivative for a quadratic radical.
The key is to inspect the whole integrand before choosing a method. A substitution that handles the square root may make the remaining rational factor more complicated; conversely, a small algebraic rewrite can expose a familiar antiderivative immediately.
Recognizing the Completed Square
A quadratic \(ax^2+bx+c\), with \(a\ne 0\), can be written as \(a(x-h)^2+k\) by completing the square. For example, \(x^2+3x+7=(x+\frac{3}{2})^2+\frac{19}{4}\). This form makes it clear that the expression is positive for every real \(x\).
The sign and domain matter. The square root \(\sqrt{(x-h)^2+a^2}\), where \(a>0\), is defined for every real \(x\); a factor \(1/x\), however, still excludes \(x=0\). An antiderivative containing a logarithm or a reciprocal is valid only on intervals where the original integrand is defined.
When the Radicand's Derivative Is Present
If a quadratic \(Q(x)\) occurs as a power in the denominator and its derivative appears in the numerator, the substitution \(u=Q(x)\) can reduce the integral to a power of \(u\). For instance, \(Q'(x)/Q(x)^{3/2}\) becomes \(u^{-3/2}\,du\). This works because the chain rule gives \(\frac{d}{dx}Q(x)^{-1/2}=-\frac12 Q(x)^{-3/2}Q'(x)\).
Worked Example: A Derivative Matched to the Quadratic
Evaluate \(\displaystyle \int \frac{2x+3}{(x^2+3x+7)^{3/2}}\,dx\).
Set \(Q(x)=x^2+3x+7\). Completing the square gives \(Q(x)=(x+\frac32)^2+\frac{19}{4}\), so \(Q(x)>0\) for every real \(x\). Also, \(Q'(x)=2x+3\), exactly the numerator.
Let \(u=Q(x)\). Then \(du=(2x+3)\,dx\), and the integral becomes \(\int u^{-3/2}\,du\). Since \(\int u^{-3/2}\,du=-2u^{-1/2}+C\), returning to \(x\) gives \(\displaystyle -\frac{2}{\sqrt{x^2+3x+7}}+C\).
Check by differentiation: \(\displaystyle \frac{d}{dx}\left[-2Q(x)^{-1/2}\right]=(-2)(-\frac12)Q(x)^{-3/2}Q'(x)=\frac{2x+3}{(x^2+3x+7)^{3/2}}\). This is the original integrand.
A common mistake is to substitute \(u=Q(x)\) but omit the factor \(Q'(x)\). For example, treating \(\int (2x+3)Q(x)^{-3/2}\,dx\) as though it were \(\int Q(x)^{-3/2}\,dx\) loses the factor supplied by \(du\). The correction is to write \(du=(2x+3)\,dx\) before changing the integrand.
Using an Established Radical Antiderivative
Not every rational factor is the derivative of the quadratic. In that case, an algebraic identity can sometimes rewrite the integrand in terms of a known radical integral. For a positive constant \(a\), one useful antiderivative is \(\displaystyle \int \frac{dx}{x\sqrt{x^2+a^2}}=\frac{1}{a}\ln\left|\frac{x}{a+\sqrt{x^2+a^2}}\right|+C\), on intervals with \(x\ne 0\).
This formula can be checked directly. Let \(s=\sqrt{x^2+a^2}\). Then \(s'=x/s\), and \(\displaystyle \frac{d}{dx}\ln\left|\frac{x}{a+s}\right|=\frac1x-\frac{x}{s(a+s)}=\frac{a}{xs}\). Multiplying by \(1/a\) gives \(1/(x\sqrt{x^2+a^2})\), as required. The absolute value in the logarithm allows the expression to be used on either interval \(x<0\) or \(x>0\); the constant of integration may differ between those intervals.
Worked Example: A Square Root Divided by Its Variable
Evaluate \(\displaystyle \int \frac{\sqrt{x^2+9}}{x}\,dx\), for \(x\ne0\).
Write \(s=\sqrt{x^2+9}\). Multiply the numerator and denominator by \(s\), using \(s^2=x^2+9\): \(\displaystyle \frac{s}{x}=\frac{s^2}{xs}=\frac{x^2+9}{xs}=\frac{x}{s}+\frac{9}{xs}\).
Thus the integral separates into \(\displaystyle \int \frac{x}{\sqrt{x^2+9}}\,dx+9\int\frac{dx}{x\sqrt{x^2+9}}\). The first term is \(\sqrt{x^2+9}\), since its derivative is \(x/\sqrt{x^2+9}\). Apply the established logarithmic formula to the second term with \(a=3\). The result is \(\displaystyle \sqrt{x^2+9}+3\ln\left|\frac{x}{3+\sqrt{x^2+9}}\right|+C\).
To check, differentiate the radical term to obtain \(x/s\). The logarithm \(\ln|x/(3+s)|\) has derivative \(3/(xs)\), so its coefficient \(3\) gives \(9/(xs)\). Their sum is \(\displaystyle \frac{x}{s}+\frac{9}{xs}=\frac{x^2+9}{xs}=\frac{s}{x}\), which is the original integrand.
Combining an Algebraic Rewrite with a Substitution
Sometimes the integrand first needs an algebraic decomposition, after which different terms call for different methods. The following example combines the logarithmic radical antiderivative with a substitution. The decomposition is verified by recombining its terms, and the substitution is checked by differentiating the resulting expression.
Worked Example: A Rational Factor and Two Radical Methods
Evaluate \(\displaystyle \int \frac{\sqrt{x^2+4}}{x(x^2+1)}\,dx\), for \(x\ne0\).
Let \(s=\sqrt{x^2+4}\). Multiply by \(s/s\) and use \(s^2=x^2+4\): \(\displaystyle \frac{s}{x(x^2+1)}=\frac{x^2+4}{x(x^2+1)s}\). The rational factor decomposes as \(\displaystyle \frac{x^2+4}{x(x^2+1)}=\frac{4}{x}-\frac{3x}{x^2+1}\), because \(4(x^2+1)-3x^2=x^2+4\). Therefore the integral is \(\displaystyle 4\int\frac{dx}{xs}-3\int\frac{x\,dx}{(x^2+1)s}\).
For the first integral, use the established formula with \(a=2\): \(\displaystyle \int\frac{dx}{xs}=\frac12\ln\left|\frac{x}{2+s}\right|+C\). For the second, set \(u=s=\sqrt{x^2+4}\). Then \(du=(x/u)\,dx\), so \(x\,dx=u\,du\). Also \(x^2+1=u^2-3\). It follows that \(\displaystyle \int\frac{x\,dx}{(x^2+1)s}=\int\frac{du}{u^2-3}\).
Put \(b=\sqrt3\). Since \(u^2-3=(u-b)(u+b)\), \(\displaystyle \frac{1}{u^2-3}=\frac{1}{2b}\left(\frac{1}{u-b}-\frac{1}{u+b}\right)\). Integration therefore gives \(\displaystyle \int\frac{du}{u^2-3}=\frac{1}{2\sqrt3}\ln\left|\frac{u-\sqrt3}{u+\sqrt3}\right|+C\). Here \(u=s\ge2>\sqrt3\), so neither logarithm's ratio vanishes or changes sign.
Combining both terms, an antiderivative is \(\displaystyle 2\ln\left|\frac{x}{2+\sqrt{x^2+4}}\right|-\frac{\sqrt3}{2}\ln\left|\frac{\sqrt{x^2+4}-\sqrt3}{\sqrt{x^2+4}+\sqrt3}\right|+C\).
Check by differentiation. The first term differentiates to \(4/(xs)\), by the radical logarithm formula. For the second logarithm, let \(u=s\); its logarithm has derivative \(2\sqrt3\,u'/(u^2-3)\), where \(u'=x/u\). Including the coefficient \(-\sqrt3/2\), this contributes \(\displaystyle -\frac{3x}{u(u^2-3)}=-\frac{3x}{s(x^2+1)}\). The total derivative is \(\displaystyle \frac{4}{xs}-\frac{3x}{s(x^2+1)} =\frac{4(x^2+1)-3x^2}{xs(x^2+1)} =\frac{x^2+4}{xs(x^2+1)} =\frac{s}{x(x^2+1)}\), the original integrand.
A Reliable Choice of Method
These examples illustrate a practical order of operations. Do not begin by forcing a substitution. First identify the quadratic and its derivative; then check whether a simple algebraic rewrite exposes a known antiderivative. Keep track of restrictions such as \(x\ne0\) throughout, and verify the final expression by differentiation.
Use the completed form to understand where the radical is real and whether it is strictly positive.
If the derivative of the quadratic is present as a factor, test the substitution \(u=Q(x)\).
Multiplying by the radical divided by itself can turn a square root into a rational expression over that radical.
Separate terms that match an established radical antiderivative from terms suited to a substitution.
Differentiate the proposed antiderivative and compare it with the original integrand on its domain.
The algebraic rewrite is valid only where the factor used to multiply and divide is nonzero. For a positive completed-square radicand, its square root is positive, so this step introduces no new exclusions. Other factors, such as \(x\) in a denominator, retain their original restrictions.
Check Your Understanding
For each question, identify the domain restrictions and explain which step makes the integral manageable.
- For \(\displaystyle \int\frac{2x}{(x^2+5)^{3/2}}\,dx\), what substitution matches the numerator, and what is the resulting antiderivative?
- Why is \(\displaystyle \int\frac{\sqrt{x^2+16}}{x}\,dx\) undefined at \(x=0\), even though its radicand is positive for every real \(x\)?
- Rewrite \(\displaystyle \frac{\sqrt{x^2+4}}{x}\) as a sum of terms involving \(1/\sqrt{x^2+4}\), and \(1/(x\sqrt{x^2+4})\).
- In the third worked example, why does \(u=\sqrt{x^2+4}\) imply \(x\,dx=u\,du\)?
- What must be checked after finding an antiderivative involving logarithms?
