What is the integral of (4x^3-2x)ln(3x) ?

The integral of (4x^3-2x)ln(3x) is x^4ln(3x)-(x^4)/4-x^2ln(3x)+(x^2)/2+C

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integral of (4x^3-2x)ln(3x)

\int (4 x^{3} - 2 x) ln (3 x) d x

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

Solution steps

\int (4 x^{3} - 2 x) ln (3 x) d x

Expand

(4 x^{3} - 2 x) ln (3 x) : 4 x^{3} ln (3 x) - 2 x ln (3 x)

= \int 4 x^{3} ln (3 x) - 2 x ln (3 x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 4 x^{3} ln (3 x) d x - \int 2 x ln (3 x) d x

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

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

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

Simplify

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

Add a constant to the solution

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

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What is the integral of (4x^3-2x)ln(3x) ?
The integral of (4x^3-2x)ln(3x) is x^4ln(3x)-(x^4)/4-x^2ln(3x)+(x^2)/2+C