Question

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

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

eq - 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:

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