integrate $int sqrt{6x - 1} , dx$ using a table of integrals\ntable of integrals\n$\\boxed{\\quad}+c$

Explanation:

Step1: Let (u = 6x - 1)

Then (du=6dx), and (dx=\frac{1}{6}du).

Step2: Rewrite the integral

(\int\sqrt{6x - 1}dx=\frac{1}{6}\int\sqrt{u}du).

Step3: Use the power - rule for integration (\int u^n du=\frac{u^{n + 1}}{n+1}+C) ((n\neq - 1))

Here (n=\frac{1}{2}), so (\frac{1}{6}\int u^{\frac{1}{2}}du=\frac{1}{6}\times\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}+C).

Step4: Simplify the expression

(\frac{1}{6}\times\frac{u^{\frac{3}{2}}}{\frac{3}{2}}+C=\frac{1}{6}\times\frac{2}{3}u^{\frac{3}{2}}+C=\frac{1}{9}u^{\frac{3}{2}}+C).

Step5: Substitute back (u = 6x - 1)

(\frac{1}{9}(6x - 1)^{\frac{3}{2}}+C).

Answer:

(\frac{1}{9}(6x - 1)^{\frac{3}{2}}+C)