evaluate $int\frac{dx}{(16 + x^{2})^{2}}$.

Explanation:

Step1: Use trigonometric substitution

Let $x = 4\tan\theta$, then $dx=4\sec^{2}\theta d\theta$. And $16 + x^{2}=16+16\tan^{2}\theta = 16\sec^{2}\theta$. $$\int\frac{dx}{(16 + x^{2})^{2}}=\int\frac{4\sec^{2}\theta d\theta}{(16\sec^{2}\theta)^{2}}$$

Step2: Simplify the integrand

$$\int\frac{4\sec^{2}\theta d\theta}{256\sec^{4}\theta}=\frac{1}{64}\int\frac{d\theta}{\sec^{2}\theta}=\frac{1}{64}\int\cos^{2}\theta d\theta$$

Step3: Use the double - angle formula

Since $\cos^{2}\theta=\frac{1 + \cos(2\theta)}{2}$, we have $\frac{1}{64}\int\cos^{2}\theta d\theta=\frac{1}{128}\int(1+\cos(2\theta))d\theta$.

Step4: Integrate term - by - term

$$\frac{1}{128}\int(1+\cos(2\theta))d\theta=\frac{1}{128}\left(\theta+\frac{1}{2}\sin(2\theta)\right)+C$$

Step5: Back - substitute

Since $x = 4\tan\theta$, then $\theta=\arctan\left(\frac{x}{4}\right)$ and $\sin(2\theta)=\frac{2\tan\theta}{1 + \tan^{2}\theta}=\frac{2\frac{x}{4}}{1+\frac{x^{2}}{16}}=\frac{8x}{16 + x^{2}}$. $$\frac{1}{128}\left(\arctan\left(\frac{x}{4}\right)+\frac{1}{2}\cdot\frac{8x}{16 + x^{2}}\right)+C=\frac{1}{128}\arctan\left(\frac{x}{4}\right)+\frac{x}{32(16 + x^{2})}+C$$

Answer:

$\frac{1}{128}\arctan\left(\frac{x}{4}\right)+\frac{x}{32(16 + x^{2})}+C$