Question

question 17 · 1 point evaluate the indefinite integral given below for (xgeq\frac{3}{4}). (int\frac{9}{xsqrt{16x^{2}-9}}dx=) provide your answer below:

Explanation:

Step1: Use substitution

Let $u = 16x^{2}-9$, then $du=32x dx$, and $x dx=\frac{1}{32}du$.

Step2: Rewrite the integral

The integral $\int\frac{9}{x\sqrt{16x^{2}-9}}dx$ becomes $\int\frac{9}{\sqrt{u}}\times\frac{1}{32}du=\frac{9}{32}\int u^{-\frac{1}{2}}du$.

Step3: Integrate using power - rule

The power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n
eq - 1$). For $n=-\frac{1}{2}$, we have $\frac{9}{32}\int u^{-\frac{1}{2}}du=\frac{9}{32}\times\frac{u^{\frac{1}{2}}}{\frac{1}{2}}+C=\frac{9}{16}\sqrt{u}+C$.

Step4: Substitute back

Substitute $u = 16x^{2}-9$ back into the result, we get $\frac{9}{16}\sqrt{16x^{2}-9}+C$.

Answer:

$\frac{9}{16}\sqrt{16x^{2}-9}+C$