How to solve ∫ 1/x² + a² dx by trigonometric substitution?

Here is the complete derivation. **Step 1: Choose the substitution.** Let $x = a \tan \theta$, where $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$. This choice is motivated by the identity $1 + \tan^2 \theta = \sec^2 \theta$. **Step 2: Compute $dx$.** $$\frac{dx}{d\theta} = a \sec^2 \theta \quad \Rightarrow \quad dx = a \sec^2 \theta \, d\theta$$ **Step 3: Rewrite the integral.** \begin{align*} \int \frac{1}{x^2 + a^2} \, dx &= \int \frac{1}{a^2 \tan^2 \theta + a^2} \cdot a \sec^2 \theta \, d\theta \\ &= \int \frac{a \sec^2 \theta}{a^2(\tan^2 \theta + 1)} \, d\theta \\ &= \int \frac{a \sec^2 \theta}{a^2 \sec^2 \theta} \, d\theta \quad \text{(using $1+\tan^2\theta = \sec^2\theta$)} \\ &= \int \frac{1}{a} \, d\theta = \frac{\theta}{a} + C \end{align*} **Step 4: Back-substitute.** Since $x = a \tan \theta$, we have $\theta = \arctan\left(\frac{x}{a}\right)$. Therefore: $$\boxed{\int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \arctan\left(\frac{x}{a}\right) + C}$$