What is the integral of 2xln(4x) ?

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

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integral of 2xln(4x)

\int 2 x ln (4 x) d x

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

Solution steps

\int 2 x ln (4 x) d x

Take the constant out:

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

= 2 \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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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