What is expand 0.5(2x-1)(3-2(2x-1)ln(2x-1)) ?

The solution to expand 0.5(2x-1)(3-2(2x-1)ln(2x-1)) is -4x^2ln(2x-1)+3x+4xln(2x-1)-ln(2x-1)-1.5

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expand 0.5(2x-1)(3-2(2x-1)ln(2x-1))

expand

0.5 (2 x - 1) (3 - 2 (2 x - 1) ln (2 x - 1))

- 4 x^{2} ln (2 x - 1) + 3 x + 4 x ln (2 x - 1) - ln (2 x - 1) - 1.5

Solution steps

0.5 (2 x - 1) (3 - 2 (2 x - 1) ln (2 x - 1))

= 0.5 (2 x - 1) (3 - 2 ln (2 x - 1) (2 x - 1))

Expand

(2 x - 1) (3 - 2 (2 x - 1) ln (2 x - 1)) : - 8 x^{2} ln (2 x - 1) + 6 x + 8 x ln (2 x - 1) - 2 ln (2 x - 1) - 3

= 0.5 (- 8 x^{2} ln (2 x - 1) + 6 x + 8 x ln (2 x - 1) - 2 ln (2 x - 1) - 3)

Expand

0.5 (- 8 x^{2} ln (2 x - 1) + 6 x + 8 x ln (2 x - 1) - 2 ln (2 x - 1) - 3) : - 4 x^{2} ln (2 x - 1) + 3 x + 4 x ln (2 x - 1) - ln (2 x - 1) - 1.5

= - 4 x^{2} ln (2 x - 1) + 3 x + 4 x ln (2 x - 1) - ln (2 x - 1) - 1.5

6 + 5 (m + 1 = 20)

\frac{( x - 1 ) ( x ^{2} + 1 )}{3 ( x + 1 )}

\frac{- ( - 1 ) ^{x - 1} - 15 + 4 x}{2}

v (3, - 2) F (3, 4)

\frac{4 ( x + h ) + 7}{3 ( x + h ) - 2}

What is expand 0.5(2x-1)(3-2(2x-1)ln(2x-1)) ?
The solution to expand 0.5(2x-1)(3-2(2x-1)ln(2x-1)) is -4x^2ln(2x-1)+3x+4xln(2x-1)-ln(2x-1)-1.5