Conquering the Nasty Surd Integrals

Integrals that involve $\sqrt{x^2\pm a^2}$ are a notorious time‑sink because they usually demand a trigonometric substitution, but the HSC syllabus actually permits you to memorise the resulting logarithmic forms. By internalising these shortcuts, you can leap straight to the answer. In this post we’ll see how to use the formulas on simple plug‑and‑play problems, on quadratics that need completing the square, and even on a differential equation that separates into these same surd integrals.

Problem Statement

Let's be real: integration can be brutal. You're in the middle of a high-pressure HSC exam, you spot an integral with a $\sqrt{x^2 \pm a^2}$ in the denominator, and your heart sinks. You know what's coming: a messy trigonometric substitution (like $x = a \tan \theta$) and a nasty $\sec \theta$ integration.

But what if I told you there's a legal cheat code built right into the syllabus?

Syllabus Secret "Students may benefit from, but are not expected to, recognise and use the results $\int \frac{dx}{\sqrt{x^2+a^2}} = \ln(x + \sqrt{x^2+a^2}) + c$ and $\int \frac{dx}{\sqrt{x^2-a^2}} = \ln(x + \sqrt{x^2-a^2}) + c$ (where $x > a$). These results can be shown using a trigonometric substitution, however, the result for $\int \sec \theta d\theta$ is required."

Translation: You are absolutely allowed to just memorize these logarithmic forms. By keeping these in your mental toolkit, you bypass the entire trig derivation. Let's look at how fast this is in practice.

Level 1: The Plug-and-Play (Easy)

The Problem:

$$\int \frac{1}{\sqrt{x^2 + 16}} \, dx$$

Level 2: The Classic 'Complete the Square' (Medium-Hard)

The Problem:

$$\int \frac{1}{\sqrt{x^2 - 6x + 5}} \, dx, \quad \text{for } x > 5$$

Level 3: The Q16 Boss Battle (Hard / Ext 2 Enrichment)

The Problem: Find the explicit function $y(x)$ that satisfies:

$$\frac{dy}{dx} = \frac{y \ln(y)}{\sqrt{x^2 - 4}}$$

given $y(2\sqrt{2}) = e^2$, for $x > 2$ and $y > 1$.

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Hints

• Instead of using a full trigonometric substitution, try to match the integrand to the standard logarithmic forms.
• If the denominator has a quadratic that isn't a perfect sum/difference of squares, try completing the square first.
• For the differential equation, separate the variables $x$ and $y$ on different sides of the equation before integrating.

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Solutions

Level 1 Solution: We can see that the integrand is exactly of the form $\frac{1}{\sqrt{x^2+a^2}}$ with $a = 4$. Match to the standard form $\int \frac{dx}{\sqrt{x^2+a^2}}$ where $a^2 = 16$. Just write the answer:

$$\int \frac{1}{\sqrt{x^2 + 16}} \, dx = \ln\left(x + \sqrt{x^2 + 16}\right) + C$$

Pro Tip: For the $x^2 + a^2$ case, $x + \sqrt{x^2+a^2}$ is always positive, so standard brackets are perfectly fine!

Level 2 Solution: The quadratic $x^2-6x+5$ is not yet a difference of squares, so we complete the square: $x^2-6x+5 = (x-3)^2 - 4$. The integral therefore becomes

$$\int \frac{1}{\sqrt{(x-3)^2 - 4}} \, dx$$

which matches the standard form $\int \frac{du}{\sqrt{u^2-a^2}}$ with $u = x-3$ and $a = 2$. Applying the logarithmic result directly gives

$$= \ln\left((x-3) + \sqrt{(x-3)^2 - 4}\right) + C$$

Expand the surd back to its original form to clean it up:

$$= \ln\left(x - 3 + \sqrt{x^2 - 6x + 5}\right) + C$$

Level 3 Solution: The differential equation is separable. Rearranging puts all $y$ terms on the left and all $x$ terms on the right:

$$\frac{dy}{y \ln y} = \frac{dx}{\sqrt{x^2 - 4}}$$

Now we integrate both sides. On the left, the substitution $u = \ln y$ gives $\int \frac{1}{u}\,du = \ln|u|$; since $y > 1$, $\ln y > 0$ and we may drop the absolute value to obtain $\ln(\ln y)$. On the right, $\int \frac{dx}{\sqrt{x^2-4}}$ matches the standard form with $a=2$, so we get $\ln(x + \sqrt{x^2-4})$. Hence

$$\ln(\ln y) = \ln\left(x + \sqrt{x^2 - 4}\right) + C$$

Sub in the initial condition ($x = 2\sqrt{2}, y = e^2$) to find $C$:

$$\ln(\ln e^2) = \ln\left(2\sqrt{2} + \sqrt{8 - 4}\right) + C$$

$$\ln(2) = \ln(2\sqrt{2} + 2) + C$$

$$C = \ln(2) - \ln(2\sqrt{2} + 2) = \ln\left(\frac{1}{\sqrt{2} + 1}\right) = \ln(\sqrt{2} - 1)$$

Now we substitute $C$ back and use log laws to combine the right‑hand side into a single logarithm:

$$\ln(\ln y) = \ln\left(x + \sqrt{x^2 - 4}\right) + \ln(\sqrt{2} - 1)$$

$$\ln(\ln y) = \ln\left((\sqrt{2} - 1)(x + \sqrt{x^2 - 4})\right)$$

Exponentiating both sides removes the outer logarithm (since the natural log is one‑to‑one):

$$\ln y = (\sqrt{2} - 1)\left(x + \sqrt{x^2 - 4}\right)$$

$$y = e^{(\sqrt{2} - 1)\left(x + \sqrt{x^2 - 4}\right)}$$

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Takeaways

• Memorizing the logarithmic integrals for $\sqrt{x^2 \pm a^2}$ is a massive time-saver and perfectly valid in HSC exams.
• Completing the square is an essential technique for reducing complex quadratics into standard integrable forms.
• Don't forget that log laws can elegantly simplify ugly constants of integration into exact, clean expressions.

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Further Readings

HSC Integrals, HSC Differential Equations