evaluate the following integral.\n int\frac{dx}{(6 - 8x)^{3}}\n int\frac{dx}{(6 - 8x)^{3}}=square

Explanation:

Step1: Use substitution

Let $u = 6-8x$, then $du=-8dx$, and $dx=-\frac{1}{8}du$.

Step2: Rewrite the integral

The integral $\int\frac{dx}{(6 - 8x)^3}$ becomes $\int\frac{-\frac{1}{8}du}{u^3}=-\frac{1}{8}\int u^{- 3}du$.

Step3: Integrate using power - rule

The power - rule for integration is $\int x^n dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$). For $n=-3$, we have $-\frac{1}{8}\int u^{-3}du=-\frac{1}{8}\times\frac{u^{-3 + 1}}{-3+1}+C$. $=-\frac{1}{8}\times\frac{u^{-2}}{-2}+C=\frac{1}{16u^{2}}+C$.

Step4: Substitute back $u = 6-8x$

The result is $\frac{1}{16(6 - 8x)^{2}}+C$.

Answer:

$\frac{1}{16(6 - 8x)^{2}}+C$