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

The integral of xln(\sqrt[3]{x}) is 1/3 (1/2 x^2ln(x)-(x^2)/4)+C

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

\frac{1}{3} (\frac{1}{2} x^{2} ln (x) - \frac{x ^{2}}{4}) + C

Solution steps

Simplify

= \int \frac{1}{3} x ln (x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int x ln (x) d x

Apply Integration By Parts

: \frac{1}{2} x^{2} ln (x) - \int \frac{x}{2} d x

= \frac{1}{3} (\frac{1}{2} x^{2} ln (x) - \int \frac{x}{2} d x)

\int \frac{x}{2} d x = \frac{x ^{2}}{4}

= \frac{1}{3} (\frac{1}{2} x^{2} ln (x) - \frac{x ^{2}}{4})

Add a constant to the solution

= \frac{1}{3} (\frac{1}{2} x^{2} ln (x) - \frac{x ^{2}}{4}) + C

f (x) = - \frac{1}{x}

f (x) = 1 + x

y^{^{'}} = 5 x - 3 x y

\int \frac{\frac{π}{6}}{0} sin (x) cos^{3} (x) d x

\frac{1}{s ^{2}}

What is the integral of xln(\sqrt[3]{x}) ?
The integral of xln(\sqrt[3]{x}) is 1/3 (1/2 x^2ln(x)-(x^2)/4)+C