What is the integral of (2-x)^2ln(4x) ?

The integral of (2-x)^2ln(4x) is 4xln(4x)-2x^2ln(4x)+1/3 x^3ln(4x)-4x+x^2-(x^3)/9+C

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integral of (2-x)^2ln(4x)

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

4 x ln (4 x) - 2 x^{2} ln (4 x) + \frac{1}{3} x^{3} ln (4 x) - 4 x + x^{2} - \frac{x ^{3}}{9} + C

Solution steps

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

Apply Integration By Parts

: ln (4 x) (4 x - 2 x^{2} + \frac{x ^{3}}{3}) - \int \frac{1}{3} (12 - 6 x + x^{2}) d x

= ln (4 x) (4 x - 2 x^{2} + \frac{x ^{3}}{3}) - \int \frac{1}{3} (12 - 6 x + x^{2}) d x

\int \frac{1}{3} (12 - 6 x + x^{2}) d x = 4 x - x^{2} + \frac{x ^{3}}{9}

= ln (4 x) (4 x - 2 x^{2} + \frac{x ^{3}}{3}) - (4 x - x^{2} + \frac{x ^{3}}{9})

Simplify

ln (4 x) (4 x - 2 x^{2} + \frac{x ^{3}}{3}) - (4 x - x^{2} + \frac{x ^{3}}{9}) : 4 x ln (4 x) - 2 x^{2} ln (4 x) + \frac{1}{3} x^{3} ln (4 x) - 4 x + x^{2} - \frac{x ^{3}}{9}

= 4 x ln (4 x) - 2 x^{2} ln (4 x) + \frac{1}{3} x^{3} ln (4 x) - 4 x + x^{2} - \frac{x ^{3}}{9}

Add a constant to the solution

= 4 x ln (4 x) - 2 x^{2} ln (4 x) + \frac{1}{3} x^{3} ln (4 x) - 4 x + x^{2} - \frac{x ^{3}}{9} + C

\frac{11}{3} t^{3} + 11 t^{2}

\frac{d}{d x} ((6 x^{2} + 6) e^{- x})

\frac{d}{d x} (e^{l n (2)})

y = 3 x^{3} - 3 x, (- 1, 0)

\frac{\partial}{\partial t} (\frac{( r + s )}{t})

What is the integral of (2-x)^2ln(4x) ?
The integral of (2-x)^2ln(4x) is 4xln(4x)-2x^2ln(4x)+1/3 x^3ln(4x)-4x+x^2-(x^3)/9+C