What is the integral of 4(x-1)ln(x-1) ?

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

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integral of 4(x-1)ln(x-1)

\int 4 (x - 1) ln (x - 1) d x

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

Solution steps

\int 4 (x - 1) ln (x - 1) d x

Take the constant out:

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

= 4 \cdot \int (x - 1) ln (x - 1) d x

Expand

(x - 1) ln (x - 1) : x ln (x - 1) - ln (x - 1)

= 4 \cdot \int x ln (x - 1) - ln (x - 1) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 4 (\int x ln (x - 1) d x - \int ln (x - 1) d x)

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

\int ln (x - 1) d x = x ln (x - 1) - ln (x - 1) - x + 1

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

Simplify

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

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

Add a constant to the solution

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

y^{^{'}} - \frac{3}{x} y = \frac{y ^{4}}{x ^{7}}

\frac{d}{d x} (4 sin^{2} (3 x))

y^{^{''}} + 10 y^{^{'}} + 25 y = e^{- 4 x} (x^{2} + 2 x - 4)

x^{^{''}} - 2 x = 0

\frac{d}{d x} (cos^{2} (e^{x}))

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