h. $int\frac{1}{(16 + x^{2})^{\frac{3}{2}}}dx$

Explanation:

Step1: Use trigonometric substitution

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

Step2: Rewrite the integral

Substitute into the integral: [ \begin{align*} \int\frac{1}{(16 + x^{2})^{\frac{3}{2}}}dx&=\int\frac{4\sec^{2}\theta}{(16\sec^{2}\theta)^{\frac{3}{2}}}d\theta\ &=\int\frac{4\sec^{2}\theta}{64\sec^{3}\theta}d\theta\ &=\frac{1}{16}\int\frac{1}{\sec\theta}d\theta\ &=\frac{1}{16}\int\cos\theta d\theta \end{align*} ]

Step3: Integrate $\cos\theta$

We know that $\int\cos\theta d\theta=\sin\theta + C$. So $\frac{1}{16}\int\cos\theta d\theta=\frac{1}{16}\sin\theta + C$.

Step4: Back - substitute

Since $x = 4\tan\theta$, then $\tan\theta=\frac{x}{4}$. We can form a right - triangle where the opposite side is $x$ and the adjacent side is $4$, and the hypotenuse is $\sqrt{16 + x^{2}}$. So $\sin\theta=\frac{x}{\sqrt{16 + x^{2}}}$. The integral becomes $\frac{x}{16\sqrt{16 + x^{2}}}+C$.

Answer:

$\frac{x}{16\sqrt{16 + x^{2}}}+C$