evaluate the integral.\n int\frac{dx}{(289 + x^{2})^{\frac{3}{2}}}=square \n(type an exact answer.)

Explanation:

Step1: Use trigonometric substitution

Let $x = 17\tan\theta$, then $dx=17\sec^{2}\theta d\theta$. And $289 + x^{2}=289+289\tan^{2}\theta = 289(1 + \tan^{2}\theta)=289\sec^{2}\theta$.

Step2: Rewrite the integral

The integral $\int\frac{dx}{(289 + x^{2})^{\frac{3}{2}}}$ becomes $\int\frac{17\sec^{2}\theta d\theta}{(289\sec^{2}\theta)^{\frac{3}{2}}}=\int\frac{17\sec^{2}\theta d\theta}{289^{\frac{3}{2}}\sec^{3}\theta}=\int\frac{17\sec^{2}\theta d\theta}{17^{3}\sec^{3}\theta}=\frac{1}{289}\int\frac{d\theta}{\sec\theta}=\frac{1}{289}\int\cos\theta d\theta$.

Step3: Integrate $\cos\theta$

$\frac{1}{289}\int\cos\theta d\theta=\frac{1}{289}\sin\theta + C$.

Step4: Express $\sin\theta$ in terms of $x$

Since $x = 17\tan\theta$, we have $\tan\theta=\frac{x}{17}$. Using the right - triangle relationship $\tan\theta=\frac{x}{17}=\frac{opposite}{adjacent}$, the hypotenuse $r=\sqrt{x^{2}+289}$. So $\sin\theta=\frac{x}{\sqrt{x^{2}+289}}$.

Answer:

$\frac{x}{289\sqrt{x^{2}+289}}+C$