A Reciprocal Weight That Changes the Answer
The integral \(\int\frac{dx}{x\sqrt{a^2+x^2}}\) is the tangent-substitution counterpart of the arcsecant integral. The extra reciprocal factor \(\frac1x\) is not harmless: with \(x=a\tan\theta\), it turns the integrand into \(\frac{1}{a}\csc\theta\), whose integral is a logarithm. The integrand is undefined at \(x=0\), so any interval of integration must stay on one side of the origin.
Deriving the Formula
Let \(x=a\tan\theta\) with \(0<\theta<\frac\pi2\). Then \(dx=a\sec^2\theta\,d\theta\) and \(\sqrt{a^2+x^2}=a\sec\theta\). Substituting,
From the triangle with legs \(a\) and \(x\) and hypotenuse \(\sqrt{a^2+x^2}\), we have \(\csc\theta=\frac{\sqrt{a^2+x^2}}{x}\) and \(\cot\theta=\frac ax\). Hence \(\csc\theta+\cot\theta=\frac{a+\sqrt{a^2+x^2}}{x}\), which gives the stated formula. For \(x<0\), the same expression holds with the absolute value, because the logarithm of the absolute value has the same derivative.
An Algebraic Cross-Check
Let \(u=\sqrt{a^2+x^2}\). Then \(x^2=u^2-a^2\) and \(x\,dx=u\,du\), so
This form is equivalent to the trigonometric one, because \((u-a)(u+a)=x^2\) implies \(\frac{u-a}{u+a}=\frac{x^2}{(u+a)^2}\), and so \(\frac{1}{2a}\ln\frac{u-a}{u+a}=\frac1a\ln\frac{|x|}{u+a}=-\frac1a\ln\frac{u+a}{|x|}\). The second route requires the partial-fraction integral \(\int\frac{du}{u^2-a^2}\), which you will study later.
Checking by Differentiation
Write the antiderivative as \(-\frac1a\bigl[\ln(a+u)-\ln|x|\bigr]\) with \(u=\sqrt{a^2+x^2}\) and \(u'=\frac xu\). The derivative is \(-\frac1a\bigl[\frac{x/u}{a+u}-\frac1x\bigr]=-\frac1a\cdot\frac{x^2-u(a+u)}{xu(a+u)}\). Since \(x^2=u^2-a^2\), the numerator becomes \(u^2-a^2-ua-u^2=-a(a+u)\), so the derivative is \(\frac{1}{xu}\), the integrand.
Worked Examples
Worked Example: Radius One
Evaluate \(\int_1^{\sqrt3}\frac{dx}{x\sqrt{1+x^2}}\).
With \(a=1\), \(F(x)=-\ln\frac{1+\sqrt{1+x^2}}{x}\). At \(x=\sqrt3\): \(-\ln\frac{1+2}{\sqrt3}=-\ln\sqrt3=-\frac12\ln3\). At \(x=1\): \(-\ln(1+\sqrt2)\). Subtracting,
The integrand falls from \(\frac{1}{\sqrt2}\approx0.707\) to \(\frac{1}{2\sqrt3}\approx0.289\), so an average of about \(0.45\) over a width \(0.73\) gives roughly \(0.33\).
Worked Example: Radius Two
Evaluate \(\int_2^{2\sqrt3}\frac{dx}{x\sqrt{4+x^2}}\).
With \(a=2\), \(F(x)=-\frac12\ln\frac{2+\sqrt{4+x^2}}{x}\). At \(x=2\sqrt3\): \(\sqrt{16}=4\), so \(-\frac12\ln\frac{6}{2\sqrt3}=-\frac12\ln\sqrt3=-\frac14\ln3\). At \(x=2\): \(\sqrt8=2\sqrt2\), so \(-\frac12\ln\frac{2+2\sqrt2}{2}=-\frac12\ln(1+\sqrt2)\). Hence
The result is exactly half of the radius-one example, since scaling \(x\mapsto2x\) leaves \(\frac{dx}{x}\) unchanged while \(\sqrt{4+x^2}\) doubles, halving the integrand.
Worked Example: Radius Three, Indefinite
Find \(\int\frac{dx}{x\sqrt{9+x^2}}\).
Apply the formula with \(a=3\):
The coefficient \(-\frac13\) is \(-\frac1a\), and the absolute value allows negative \(x\).
Worked Example: A Negative Interval
Evaluate \(\int_{-\sqrt3}^{-1}\frac{dx}{x\sqrt{1+x^2}}\).
The antiderivative with absolute value is \(-\ln\frac{1+\sqrt{1+x^2}}{|x|}\). At \(x=-1\): \(-\ln(1+\sqrt2)\). At \(x=-\sqrt3\): \(-\frac12\ln3\). The integral equals \(F(-1)-F(-\sqrt3)\):
The integrand is odd, so the integral over the negative interval is the negative of the integral over the mirror interval \([1,\sqrt3]\). The sign confirms that the integrand is negative for \(x<0\).
Intervals Containing Zero
As \(x\to0\), the integrand behaves like \(\frac{1}{ax}\), and the integral diverges logarithmically. The formula reflects this: \(-\frac1a\ln\frac{a+\sqrt{a^2+x^2}}{|x|}\) tends to \(-\infty\) as \(x\to0^+\). Consequently, \(\int_0^1\frac{dx}{x\sqrt{1+x^2}}\) diverges, and an interval such as \([-1,1]\) cannot be integrated across the singularity, even by symmetry. On the other side, the improper integral to infinity converges: \(\int_1^\infty\frac{dx}{x\sqrt{1+x^2}}=\ln(1+\sqrt2)\approx0.881\), because the integrand decays like \(x^{-2}\).
A Reliable Procedure
Make sure it does not contain \(x=0\).
Set \(x=a\tan\theta\), so the integrand becomes \(\frac1a\csc\theta\,d\theta\).
Use \(\int\csc\theta\,d\theta=-\ln|\csc\theta+\cot\theta|\).
Use \(\csc\theta=\frac{\sqrt{a^2+x^2}}{x}\) and \(\cot\theta=\frac ax\).
Differentiate, or compare with \(\frac{1}{2a}\ln\left|\frac{u-a}{u+a}\right|\).
Common Mistakes
The leading error is dropping the minus sign in front of \(\frac1a\); the antiderivative must be increasing where the integrand is positive, and \(-\ln\frac{a+\sqrt{a^2+x^2}}{x}\) indeed increases for \(x>0\). Another is to forget the absolute value and then be unable to handle negative \(x\). A third is to invert the quotient inside the logarithm, writing \(\ln\frac{x}{a+\sqrt{a^2+x^2}}\) together with the leading minus sign. That double counts the sign and reverses the answer.
Quick Reference
| Integrand | Antiderivative |
|---|---|
| \(\frac{1}{x\sqrt{1+x^2}}\) | \(-\ln\frac{1+\sqrt{1+x^2}}{|x|}\) |
| \(\frac{1}{x\sqrt{9+x^2}}\) | \(-\frac13\ln\frac{3+\sqrt{9+x^2}}{|x|}\) |
| \(\frac{1}{x\sqrt{4+9x^2}}\) | \(-\frac12\ln\frac{2+\sqrt{4+9x^2}}{3|x|}\) |
| \(\frac{1}{x\sqrt{a^2+x^2}}\) | \(-\frac1a\ln\frac{a+\sqrt{a^2+x^2}}{|x|}\) |
A Numerical Spot Check
Check \(\int_1^{\sqrt3}\frac{dx}{x\sqrt{1+x^2}}\approx0.332\) with the trapezoid rule on one interval: the heights are \(\frac{1}{\sqrt2}=0.7071\) at \(x=1\) and \(\frac{1}{2\sqrt3}=0.2887\) at \(x=\sqrt3\), so the estimate is \(0.7321\cdot\frac{0.7071+0.2887}{2}\approx0.3645\). It exceeds \(0.332\) because the integrand is convex and decreasing, and the discrepancy of about ten percent is typical of one trapezoid.
Practice Problems with Answers
Try each problem before reading the answer. Every answer below was checked by differentiation or by numerical integration.
- Evaluate \(\int_1^{\sqrt3}\frac{dx}{x\sqrt{1+x^2}}\). Hint: use \(F(x)=-\ln\frac{1+\sqrt{1+x^2}}{x}\). Answer: \(\ln(1+\sqrt2)-\frac12\ln3\)
- Evaluate \(\int_2^{2\sqrt3}\frac{dx}{x\sqrt{4+x^2}}\). Hint: here \(a=2\). Answer: \(\frac12\ln(1+\sqrt2)-\frac14\ln3\)
- Find an antiderivative of \(\frac{1}{x\sqrt{9+x^2}}\) for \(x>0\). Hint: use \(a=3\). Answer: \(-\frac13\ln\frac{3+\sqrt{9+x^2}}{x}\)
- Evaluate \(\int_{-\sqrt3}^{-1}\frac{dx}{x\sqrt{1+x^2}}\). Hint: the integrand is negative for \(x<0\). Answer: \(\frac12\ln3-\ln(1+\sqrt2)\)
Worked Example: A Scaled Radical
Find \(\int\frac{dx}{x\sqrt{4+9x^2}}\) for \(x>0\).
Let \(u=3x\). Then \(\frac{dx}{x}=\frac{du}{u}\), and \(\sqrt{4+9x^2}=\sqrt{4+u^2}\). The integral becomes \(\int\frac{du}{u\sqrt{4+u^2}}=-\frac12\ln\frac{2+\sqrt{4+u^2}}{u}\), so
The factor \(\frac1x\) is scale-invariant, so no extra constant appears from \(dx\); only the argument of the logarithm changes.
Hyperbolic Form of the Answer
Since \(\frac{a+\sqrt{a^2+x^2}}{x}=\frac{a}{x}+\sqrt{\frac{a^2}{x^2}+1}\), the logarithm equals \(\operatorname{arcsinh}\frac{a}{x}\) for \(x>0\). The antiderivative can therefore be written as \(-\frac1a\operatorname{arcsinh}\frac{a}{|x|}\), the same function as before. This form makes the behavior transparent: as \(x\to\infty\), \(\frac a{|x|}\to0\) and the antiderivative tends to \(0\) from below, and as \(x\to0^+\) it tends to \(-\infty\).
Interpreting the Result
The antiderivative is negative for all \(x>0\) and increasing, approaching zero as \(x\to\infty\). Hence the area to infinity is \(0-F(x_0)=\frac1a\operatorname{arcsinh}\frac{a}{x_0}\), which is positive and finite. The integral from a point \(x_0\) to infinity measures how much of the tail lies beyond \(x_0\), and it shrinks like \(\frac{1}{x_0}\) for large \(x_0\), since \(\operatorname{arcsinh}t\approx t\) for small \(t\). This matches the integrand's decay like \(\frac{1}{x^2}\) at infinity.
Comparison with the Secant Case
| Feature | \(\frac{1}{x\sqrt{x^2-a^2}}\) | \(\frac{1}{x\sqrt{a^2+x^2}}\) |
|---|---|---|
| Substitution | \(x=a\sec\theta\) | \(x=a\tan\theta\) |
| Leftover integral | \(\frac1a\int d\theta\) | \(\frac1a\int\csc\theta\,d\theta\) |
| Type of answer | inverse trigonometric | logarithmic |
| Domain | \(|x|>a\) | \(x\ne0\) |
The difference arises because the tangent substitution leaves a cosecant, while the secant substitution leaves a constant. The domain also differs: the tangent case is defined for all nonzero \(x\), so intervals may be arbitrarily close to zero from either side.
Where the Absolute Value Comes From
The antiderivative of \(\csc\theta\) is \(-\ln|\csc\theta+\cot\theta|\), and the absolute value is part of that standard formula. When \(x>0\), \(\theta\) lies in \((0,\frac\pi2)\) and both \(\csc\theta\) and \(\cot\theta\) are positive, so the absolute value can be dropped. When \(x<0\), \(\theta\) lies in \((-\frac\pi2,0)\), where \(\csc\theta\) and \(\cot\theta\) are both negative, and their sum is negative. The absolute value then makes the argument positive. Writing the answer in terms of \(|x|\) accounts for both cases at once, because the quotient \(\frac{a+\sqrt{a^2+x^2}}{|x|}\) is always positive.
A useful habit is to state the interval before you start. If the interval is \([1,2]\), say so and drop the absolute value. If the interval is negative, keep it or use the odd symmetry of the integrand to reflect the problem onto a positive interval. This avoids sign errors that are easy to make when evaluating logarithms at negative endpoints.
Where This Fits in the Course
This tutorial belongs to the module “Trigonometric Substitution”. The tutorial just before it is “Polynomial Factors Over √(a^2 + x^2),” and the one that follows is “Definite Integrals With a^2 + x^2 Radicals.” It is the tangent-substitution partner of the arcsecant integral, with a logarithm in place of an angle. The next tutorial puts these radicals into definite integrals, where the bounds must be transformed along with the variable and the original domain must be checked at the end. The habits practiced here, stating the interval first and tracking signs carefully, make that step much easier.
Check Your Understanding
Use the derivation and examples above to answer these questions.
- Why does \(x=a\tan\theta\) turn the integrand into \(\frac1a\csc\theta\)?
- Why must the interval of integration avoid \(x=0\)?
- Evaluate \(\int_1^{\sqrt3}\frac{dx}{x\sqrt{1+x^2}}\).
- Why is the antiderivative written with \(|x|\)?
- Differentiate the antiderivative and confirm the integrand.
