What is the integral of 3xln(4x) ?

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

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integral of 3xln(4x)

\int 3 x ln (4 x) d x

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

Solution steps

\int 3 x ln (4 x) d x

Take the constant out:

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

= 3 \cdot \int x ln (4 x) d x

Apply Integration By Parts

: \frac{1}{2} x^{2} ln (4 x) - \int \frac{x}{2} d x

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

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

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

Add a constant to the solution

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

f (x) = (3 - 5 x)^{9}

\frac{d}{d x} (cos^{2} ((a - x)^{3}))

t^{2} cos (8 t)

\frac{d}{d x} (x^{3} - 8 x^{2} + 31 x)

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