What is the integral of 6xln(3x) ?

The integral of 6xln(3x) is 6(1/2 x^2ln(3x)-(x^2)/4)+C

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integral of 6xln(3x)

\int 6 x ln (3 x) d x

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

Solution steps

\int 6 x ln (3 x) d x

Take the constant out:

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

= 6 \cdot \int x ln (3 x) d x

Apply Integration By Parts

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

= 6 (\frac{1}{2} x^{2} ln (3 x) - \int \frac{x}{2} d x)

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

= 6 (\frac{1}{2} x^{2} ln (3 x) - \frac{x ^{2}}{4})

Add a constant to the solution

= 6 (\frac{1}{2} x^{2} ln (3 x) - \frac{x ^{2}}{4}) + C

\frac{d}{d x} (- 3 x - 2)

\int x (x + 2) d x

\int \frac{e ^{x} e ^{x} - 1}{e ^{x} + 3} d x

\frac{d y}{d x} = \frac{1}{x ^{2} + 1} - \frac{1}{1 - x ^{2}}

n = 1 \sum \infty (\frac{1}{8})^{n}

What is the integral of 6xln(3x) ?
The integral of 6xln(3x) is 6(1/2 x^2ln(3x)-(x^2)/4)+C