Derive and apply the standard integral ∫dx/(x² + a²) = (1/a)tan⁻¹(x/a).

∫dx/(x² + a²) — Standard Integral Form

Question

Evaluate:

\int \frac{dx}{x^2 + a^2}

State the result and derive it from first principles using a trigonometric substitution.

Solution — Step by Step

The denominator is a sum of squares — no factoring will help here. The trick is to see that $x^2 + a^2$ can be rewritten as $a^2\left(\frac{x^2}{a^2} + 1\right)$ , which hints at a substitution involving $\tan\theta$ .

Let $x = a\tan\theta$ . Then:

dx = a\sec^2\theta \, d\theta

Substitute into the denominator:

x^2 + a^2 = a^2\tan^2\theta + a^2 = a^2(\tan^2\theta + 1) = a^2\sec^2\theta

The integral becomes:

\int \frac{a\sec^2\theta \, d\theta}{a^2\sec^2\theta} = \int \frac{1}{a} \, d\theta = \frac{\theta}{a} + C

The $\sec^2\theta$ terms cancel cleanly — that’s exactly why we chose $x = a\tan\theta$ .

Since $x = a\tan\theta$ , we have $\tan\theta = \frac{x}{a}$ , so $\theta = \tan^{-1}\!\left(\frac{x}{a}\right)$ .

\int \frac{dx}{x^2 + a^2} = \frac{1}{a}\tan^{-1}\!\left(\frac{x}{a}\right) + C

Final Answer:

\boxed{\int \frac{dx}{x^2 + a^2} = \frac{1}{a}\tan^{-1}\!\left(\frac{x}{a}\right) + C}

Why This Works

The substitution $x = a\tan\theta$ is not magic — it’s the natural choice because $\tan^2\theta + 1 = \sec^2\theta$ is the only Pythagorean identity that converts a sum of squares into a perfect square in the denominator. If we’d tried $x = a\sin\theta$ , we’d get $a^2(1 - \sin^2\theta) = a^2\cos^2\theta$ , which handles $a^2 - x^2$ , not $x^2 + a^2$ .

The $\frac{1}{a}$ factor in the answer comes from factoring $a^2$ out of the denominator. Students who forget this factor lose a mark every single time — more on that below.

This result is a standard formula listed in NCERT Part 2, and in JEE Main it appears in two forms: the pure integral and as a step inside a larger problem (partial fractions, completing the square). Memorise the formula, but derive it once so you understand where it comes from.

Alternative Method

We can also verify this by differentiating the result directly — a useful check in exams.

Let $F(x) = \frac{1}{a}\tan^{-1}\!\left(\frac{x}{a}\right)$ . Differentiate using the chain rule:

F'(x) = \frac{1}{a} \cdot \frac{1}{1 + \left(\frac{x}{a}\right)^2} \cdot \frac{1}{a} = \frac{1}{a^2} \cdot \frac{a^2}{a^2 + x^2} = \frac{1}{x^2 + a^2}

This confirms the formula without going through the substitution — and it’s how you can verify any standard integral in 30 seconds.

In JEE Main 2023, this integral appeared inside a partial fractions problem where students had to decompose $\frac{1}{(x+1)(x^2+4)}$ . After splitting, one piece was $\frac{1}{x^2+4}$ , directly matching this formula with $a=2$ . Recognising standard forms under pressure is what separates 160+ scorers.

Common Mistake

The most common error: writing the answer as $\tan^{-1}\!\left(\frac{x}{a}\right) + C$ , forgetting the $\frac{1}{a}$ coefficient entirely.

This happens because students memorise $\int \frac{dx}{1+x^2} = \tan^{-1}x + C$ (where $a=1$ ), and then apply it without adjusting for general $a$ . If the denominator is $x^2 + 9$ , the answer is $\frac{1}{3}\tan^{-1}\!\left(\frac{x}{3}\right) + C$ , not $\tan^{-1}\!\left(\frac{x}{3}\right) + C$ .

Quick check: differentiate your answer. If you get back $\frac{1}{x^2+a^2}$ , you’re right.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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