What is the integral of 16xln(6x) ?

The integral of 16xln(6x) is 16(1/2 x^2ln(6x)-(x^2)/4)+C

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integral of 16xln(6x)

\int 16 x ln (6 x) d x

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

Solution steps

\int 16 x ln (6 x) d x

Take the constant out:

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

= 16 \cdot \int x ln (6 x) d x

Apply Integration By Parts

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

= 16 (\frac{1}{2} x^{2} ln (6 x) - \int \frac{x}{2} d x)

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

= 16 (\frac{1}{2} x^{2} ln (6 x) - \frac{x ^{2}}{4})

Add a constant to the solution

= 16 (\frac{1}{2} x^{2} ln (6 x) - \frac{x ^{2}}{4}) + C

x y^{^{'}} + y = 8 x y^{2}

(4 x + 5)^{- 1}

n = 0 \sum \infty \frac{3}{4 ^{n}}

\frac{d}{d x} (x^{2} π)

\int \frac{( 1 + x ^{2} )}{x} d x

What is the integral of 16xln(6x) ?
The integral of 16xln(6x) is 16(1/2 x^2ln(6x)-(x^2)/4)+C