What is the integral of 8xln(4x) ?

The integral of 8xln(4x) is 8(1/2 x^2ln(4x)-(x^2)/4)+C

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integral of 8xln(4x)

\int 8 x ln (4 x) d x

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

Solution steps

\int 8 x ln (4 x) d x

Take the constant out:

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

= 8 \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

= 8 (\frac{1}{2} x^{2} ln (4 x) - \int \frac{x}{2} d x)

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

= 8 (\frac{1}{2} x^{2} ln (4 x) - \frac{x ^{2}}{4})

Add a constant to the solution

= 8 (\frac{1}{2} x^{2} ln (4 x) - \frac{x ^{2}}{4}) + C

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What is the integral of 8xln(4x) ?
The integral of 8xln(4x) is 8(1/2 x^2ln(4x)-(x^2)/4)+C