What is the integral of \sqrt[3]{x}ln(6x) ?

The integral of \sqrt[3]{x}ln(6x) is 3/4 x^{4/3}ln(6x)-9/16 x^{4/3}+C

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integral of \sqrt[3]{x}ln(6x)

\frac{3}{4} x^{\frac{4}{3}} ln (6 x) - \frac{9}{16} x^{\frac{4}{3}} + C

Solution steps

Apply Integration By Parts

: \frac{3}{4} x^{\frac{4}{3}} ln (6 x) - \int \frac{3}{4} x^{\frac{1}{3}} d x

= \frac{3}{4} x^{\frac{4}{3}} ln (6 x) - \int \frac{3}{4} x^{\frac{1}{3}} d x

\int \frac{3}{4} x^{\frac{1}{3}} d x = \frac{9}{16} x^{\frac{4}{3}}

= \frac{3}{4} x^{\frac{4}{3}} ln (6 x) - \frac{9}{16} x^{\frac{4}{3}}

Add a constant to the solution

= \frac{3}{4} x^{\frac{4}{3}} ln (6 x) - \frac{9}{16} x^{\frac{4}{3}} + C

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What is the integral of \sqrt[3]{x}ln(6x) ?
The integral of \sqrt[3]{x}ln(6x) is 3/4 x^{4/3}ln(6x)-9/16 x^{4/3}+C