What is the integral of 16xln(3x) ?

The integral of 16xln(3x) is 16(1/2 x^2ln(3x)-(x^2)/4)+C

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integral of 16xln(3x)

\int 16 x ln (3 x) d x

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

Solution steps

\int 16 x ln (3 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 (3 x) d x

Apply Integration By Parts

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

= 16 (\frac{1}{2} x^{2} ln (3 x) - \int \frac{x}{2} d x)

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

= 16 (\frac{1}{2} x^{2} ln (3 x) - \frac{x ^{2}}{4})

Add a constant to the solution

= 16 (\frac{1}{2} x^{2} ln (3 x) - \frac{x ^{2}}{4}) + C

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

x = - 2, x = 1, y = 2 x^{2} + 12, y = 0

f (x) = 5 x + 11, (5, 6)

f (x) = \frac{3}{x ^{2}}

f (x) = 3 (x^{3} - x^{2})

What is the integral of 16xln(3x) ?
The integral of 16xln(3x) is 16(1/2 x^2ln(3x)-(x^2)/4)+C