evaluate ∫ dx / (x²√(49 - x²))

# Explanation: ## Step1: Use trig - substitution Let $x = 7\sin\theta$, then $dx=7\cos\theta d\theta$. And $\sqrt{49 - x^{2}}=\sqrt{49-49\sin^{2}\theta}=7\cos\theta$, $x^{2}=49\sin^{2}\theta$. The integral $\int\frac{dx}{x^{2}\sqrt{49 - x^{2}}}$ becomes $\int\frac{7\cos\theta d\theta}{49\sin^{2}\theta\times7\cos\theta}=\frac{1}{49}\int\frac{d\theta}{\sin^{2}\theta}$. ## Step2: Recall the integral formula We know that $\int\frac{d\theta}{\sin^{2}\theta}=-\cot\theta + C$. So $\frac{1}{49}\int\frac{d\theta}{\sin^{2}\theta}=-\frac{1}{49}\cot\theta + C$. ## Step3: Express $\cot\theta$ in terms of $x$ Since $x = 7\sin\theta$, then $\sin\theta=\frac{x}{7}$, and using the right - triangle relationship $\cot\theta=\frac{\sqrt{49 - x^{2}}}{x}$. So the integral $=-\frac{\sqrt{49 - x^{2}}}{49x}+C$. # Answer: $-\frac{\sqrt{49 - x^{2}}}{49x}+C$