Why This Reciprocal Radical Has a Logarithmic Antiderivative
The integral of \(1/\sqrt{a^2+x^2}\) is a useful example of how trigonometric substitution turns a radical into a trigonometric expression that can be integrated directly. The final answer is logarithmic, but the substitution and its inverse must be handled with care: the scale factor \(a\) appears in both \(x\) and \(dx\), and the chosen angle range determines the sign of the square root.
Throughout, assume \(a>0\) is a constant and \(x\in\mathbb{R}\). This assumption ensures \(a^2+x^2>0\) for every real \(x\), so the integrand is defined on the whole real line. The earlier material on tangent substitution uses \(x=a\tan\theta\) for this type of radicand. Here we carry that substitution through the reciprocal radical and return the result to \(x\).
Deriving the Formula by Tangent Substitution
Set \(x=a\tan\theta\), choosing \(-\pi/2<\theta<\pi/2\). On this interval, \(\cos\theta>0\), and therefore \(\sec\theta>0\). The angle range matters because it gives a definite sign when simplifying the square root.
Let \(x=a\tan\theta\). Differentiating gives \(dx=a\sec^2\theta\,d\theta\).
Since \(1+\tan^2\theta=\sec^2\theta\) and \(\sec\theta>0\), \(\sqrt{a^2+x^2}=a\sec\theta\).
The factors of \(a\) and one factor of \(\sec\theta\) cancel, leaving \(\int\sec\theta\,d\theta\).
Use \(\tan\theta=x/a\) and \(\sec\theta=\sqrt{a^2+x^2}/a\) in the logarithmic result.
The substitution reduces the integral as follows:
To integrate \(\sec\theta\), multiply the integrand by \((\sec\theta+\tan\theta)/(\sec\theta+\tan\theta)\). On the chosen angle interval, \(\sec\theta+\tan\theta>0\), so the logarithm is well-defined; the absolute-value form also expresses the standard antiderivative.
The last equality follows from the chain rule: the derivative of \(\sec\theta+\tan\theta\) is \(\sec\theta\tan\theta+\sec^2\theta\), exactly the numerator in the preceding fraction. Now substitute back:
The final step absorbs \(-\ln a\), which is constant with respect to \(x\), into the arbitrary constant of integration. This explains why equivalent answers may show a factor of \(1/a\) inside the logarithm or may omit it.
Why the Logarithm Is Defined for Every Real \(x\)
For \(a>0\), the square root is strictly greater than \(|x|\), because \[ a^2+x^2>x^2 \] and both square roots are nonnegative. Hence \[ \sqrt{a^2+x^2}>|x|\ge -x. \] Adding \(x\) to the first and last terms gives \(x+\sqrt{a^2+x^2}>0\). Thus the logarithm’s argument is positive on all of \(\mathbb{R}\). Writing an absolute value is conventional, but it is not needed to make this particular logarithm defined.
This also clarifies why the condition \(a>0\) is convenient. If a problem gives a nonzero real constant that is not known to be positive, the radicand depends on its square, and the positive scale in the substitution is \(|a|\). One may state the formula using \(\sqrt{a^2}\) or first replace the parameter by its positive magnitude. The case \(a=0\) is different: the integrand \(1/\sqrt{x^2}=1/|x|\) is undefined at \(x=0\), so it does not have an antiderivative across the whole real line.
Worked Examples
Worked Example: Integrate \(1/\sqrt{9+x^2}\)
Here \(a=3\), which is positive. Use \(x=3\tan\theta\). Then \(dx=3\sec^2\theta\,d\theta\), and, with \(-\pi/2<\theta<\pi/2\), \[ \sqrt{9+x^2} =\sqrt{9+9\tan^2\theta} =3\sqrt{1+\tan^2\theta} =3\sec\theta. \] Therefore, \[ \begin{aligned} \int\frac{dx}{\sqrt{9+x^2}} &=\int\frac{3\sec^2\theta}{3\sec\theta}\,d\theta\\ &=\int\sec\theta\,d\theta\\ &=\ln|\sec\theta+\tan\theta|+C\\ &=\ln\left|\frac{\sqrt{9+x^2}}{3}+\frac{x}{3}\right|+C\\ &=\ln|x+\sqrt{9+x^2}|+C_1. \end{aligned} \] Check by differentiating. Let \(S=\sqrt{9+x^2}\), so \(S'=x/S\). Then \[ \frac{d}{dx}\ln(x+S) =\frac{1+x/S}{x+S} =\frac{(S+x)/S}{x+S} =\frac{1}{S} =\frac{1}{\sqrt{9+x^2}}. \] The logarithm’s argument is positive, so the absolute value does not affect this calculation.
Worked Example: Integrate \(1/\sqrt{25+x^2}\) and Keep the Scale Clear
Take \(a=5\), and set \(x=5\tan\theta\). The differential and radical become \[ dx=5\sec^2\theta\,d\theta, \qquad \sqrt{25+x^2} =\sqrt{25(1+\tan^2\theta)} =5\sec\theta. \] Substituting both expressions gives \[ \begin{aligned} \int\frac{dx}{\sqrt{25+x^2}} &=\int\frac{5\sec^2\theta}{5\sec\theta}\,d\theta\\ &=\int\sec\theta\,d\theta\\ &=\ln|\sec\theta+\tan\theta|+C. \end{aligned} \] Since \(\tan\theta=x/5\) and \(\sec\theta=\sqrt{25+x^2}/5\), \[ \begin{aligned} \int\frac{dx}{\sqrt{25+x^2}} &=\ln\left|\frac{x+\sqrt{25+x^2}}{5}\right|+C\\ &=\ln|x+\sqrt{25+x^2}|+C_1. \end{aligned} \] For a derivative check, put \(S=\sqrt{25+x^2}\). Then \(S'=x/S\), and \[ \frac{d}{dx}\ln(x+S) =\frac{1+x/S}{x+S} =\frac{S+x}{S(x+S)} =\frac{1}{S} =\frac{1}{\sqrt{25+x^2}}. \] The factor \(1/5\) in the logarithm changes the expression only by the constant \(-\ln 5\). It does not change the derivative.
Worked Example: Integrate \(1/\sqrt{4+x^2}\) and Check at a Negative Input
For \(a=2\), the tangent substitution is \(x=2\tan\theta\). It gives \(dx=2\sec^2\theta\,d\theta\) and \(\sqrt{4+x^2}=2\sec\theta\) on \(-\pi/2<\theta<\pi/2\). Thus \[ \begin{aligned} \int\frac{dx}{\sqrt{4+x^2}} &=\int\frac{2\sec^2\theta}{2\sec\theta}\,d\theta\\ &=\ln|\sec\theta+\tan\theta|+C\\ &=\ln\left|\frac{\sqrt{4+x^2}+x}{2}\right|+C\\ &=\ln(x+\sqrt{4+x^2})+C_1. \end{aligned} \] The last logarithm is valid without absolute values: \(\sqrt{4+x^2}>|x|\), including when \(x<0\). For a specific check at \(x=-1\), the integrand has value \[ \frac{1}{\sqrt{4+(-1)^2}}=\frac{1}{\sqrt{5}}. \] The derivative of the proposed antiderivative is \((1+x/\sqrt{4+x^2})/(x+\sqrt{4+x^2})\). At \(x=-1\), this is \[ \frac{1-1/\sqrt{5}}{-1+\sqrt{5}} =\frac{(\sqrt{5}-1)/\sqrt{5}}{\sqrt{5}-1} =\frac{1}{\sqrt{5}}, \] which agrees with the integrand. This check also confirms that the formula behaves correctly for a negative input.
A General Derivative Check
The examples verify particular parameter values. The same check proves that the formula works for every \(a>0\) and every real \(x\). Define \(S(x)=\sqrt{a^2+x^2}\). Since \(S>0\), differentiation gives \(S'(x)=x/S(x)\). Also, as shown above, \(x+S(x)>0\), so the derivative of its logarithm is valid everywhere.
This derivative check is an efficient way to confirm a result obtained by substitution. The substitution explains how the logarithm arises; differentiation independently verifies that converting back to \(x\) produced the right expression.
Common Pitfall: Losing the Scale Factor
A frequent error is to set \(x=a\tan\theta\) but write \(dx=\sec^2\theta\,d\theta\). That differential is missing a factor of \(a\). Differentiating the substitution shows exactly what is required:
For example, with \(a=3\), the incorrect differential would lead to \[ \frac{\sec^2\theta\,d\theta}{3\sec\theta} =\frac{1}{3}\sec\theta\,d\theta, \] and hence to a result with an erroneous factor of \(1/3\). The correct calculation is \[ \frac{3\sec^2\theta\,d\theta}{3\sec\theta} =\sec\theta\,d\theta. \] The factor of \(a\) in \(dx\) cancels the factor of \(a\) in the radical. When checking a trigonometric substitution, write both \(dx\) and the transformed radical explicitly before simplifying.
When This Formula Is Useful
The result provides an antiderivative for a reciprocal quadratic radical that remains real and positive for every real \(x\). Its domain is especially simple compared with forms whose radicands vanish at finite endpoints. It is also a reminder that logarithms often appear after integrating a trigonometric expression: here, the secant integral becomes a logarithm, and the inverse substitution turns its argument into \(x+\sqrt{a^2+x^2}\).
A useful final check is to compare the derivative, not merely the appearance, of two proposed answers. For example, \(\ln((x+\sqrt{a^2+x^2})/a)\) and \(\ln(x+\sqrt{a^2+x^2})\) differ by the constant \(-\ln a\) when \(a>0\). They are therefore both valid antiderivatives. The constant \(C\) accounts for this difference.
Check Your Understanding
Answer these questions by using the substitution and verification steps from the derivation.
- For \(a>0\), what angle interval makes \(\sec\theta\) positive when \(x=a\tan\theta\)?
- What is \(dx\) under the substitution \(x=a\tan\theta\), and which factor cancels against the radical?
- Why is \(x+\sqrt{a^2+x^2}\) positive for every real \(x\)?
- Differentiate \(\ln(x+\sqrt{a^2+x^2})\) to verify the general antiderivative.
- Why do the answers with and without a factor \(1/a\) inside the logarithm differ only by a constant?
