use a substitution of the form u = ax + b to evaluate the following indefinite integral.\n intsqrt{9x + 2}…

Explanation:

Step1: Set the substitution

Let $u = 9x+2$, then $du=9dx$, and $dx=\frac{1}{9}du$.

Step2: Rewrite the integral

The integral $\int\sqrt{9x + 2}dx$ becomes $\int\sqrt{u}\cdot\frac{1}{9}du=\frac{1}{9}\int u^{\frac{1}{2}}du$.

Step3: Integrate using power - rule

The power - rule for integration is $\int x^n dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$). So, $\frac{1}{9}\int u^{\frac{1}{2}}du=\frac{1}{9}\cdot\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}+C=\frac{1}{9}\cdot\frac{u^{\frac{3}{2}}}{\frac{3}{2}}+C=\frac{2}{27}u^{\frac{3}{2}}+C$.

Step4: Substitute back

Substitute $u = 9x+2$ back into the result. We get $\frac{2}{27}(9x + 2)^{\frac{3}{2}}+C$.

Answer:

$\frac{2}{27}(9x + 2)^{\frac{3}{2}}+C$