Conquering the Nasty 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 $y > 1$.

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.

Solutions

Level 1 Solution: 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: Complete the square in the denominator, then apply the formula directly (treat $(x-3)$ as your $x$): $\int \frac{1}{\sqrt{(x-3)^2 - 4}} \, dx = \ln\left((x-3) + \sqrt{(x-3)^2 - 4}\right) + C$ 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: Separate the variables and integrate both sides immediately: $\int \frac{1}{y \ln y} \, dy = \int \frac{1}{\sqrt{x^2 - 4}} \, dx$ $\ln(\ln y) = \ln\left(x + \sqrt{x^2 - 4}\right) + C$ Sub in the initial condition ($x = 2\sqrt{2}, y = e^2$) to find $C$: $\ln(\ln e^2) = \ln\left(2\sqrt{2} + \sqrt{8 - 4}\right) + C$ $\ln(2) = \ln(2\sqrt{2} + 2) + C$ $C = \ln(2) - \ln(2\sqrt{2} + 2) = \ln\left(\frac{1}{\sqrt{2} + 1}\right) = \ln(\sqrt{2} - 1)$ Substitute $C$ back and use log laws to combine the right-hand side: $\ln(\ln y) = \ln\left(x + \sqrt{x^2 - 4}\right) + \ln(\sqrt{2} - 1)$ $\ln(\ln y) = \ln\left((\sqrt{2} - 1)(x + \sqrt{x^2 - 4})\right)$ Exponentiate to drop the outer logs: $\ln y = (\sqrt{2} - 1)\left(x + \sqrt{x^2 - 4}\right)$ $y = e^{(\sqrt{2} - 1)\left(x + \sqrt{x^2 - 4}\right)}$

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.

Further Readings

• HSC Integrals: https://vumaths.com/booklets/hsc-integrals/
• HSC Differential Equations: https://vumaths.com/booklets/hsc-differential-equations/

Connect with me

If you found these cheat codes helpful, there's a lot more where that came from! Dive deeper into HSC Maths Extension 1 & 2 by checking out my YouTube channel for full video walkthroughs. You can also explore all my resources on Vu's Maths Hub or join the discussion by subscribing to my Substack. Feel free to connect with me professionally on LinkedIn!