What is the integral of 16xln(2x) ?

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

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integral of 16xln(2x)

\int 16 x ln (2 x) d x

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

Solution steps

\int 16 x ln (2 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 (2 x) d x

Apply Integration By Parts

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

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

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

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{cos ( 2 x )}{2})

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

n = 1 \sum \infty \frac{1}{9 ^{n}}

\frac{d}{d x} (\frac{x - 1}{x + 1})

\int_{0}^{1} ∣ 3 x - 2 ∣ d x

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