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

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