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

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

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integral of 4xln(4x)

\int 4 x ln (4 x) d x

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

Solution steps

\int 4 x ln (4 x) d x

Take the constant out:

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

= 4 \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

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

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

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

Simplify

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

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

Add a constant to the solution

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

ln (e^{x} x)

\frac{1}{1 - x}, (\frac{π}{4})

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

\frac{d y}{d x} = 3 x^{2} e^{- y}

\frac{1}{2 s + 2}