What is the integral of (2x-40)ln(x-20) ?

The integral of (2x-40)ln(x-20) is x^2ln(x-20)-1/2 x^2+20x-40xln(x-20)+400ln(x-20)-200+C

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integral of (2x-40)ln(x-20)

\int (2 x - 40) ln (x - 20) d x

x^{2} ln (x - 20) - \frac{1}{2} x^{2} + 20 x - 40 x ln (x - 20) + 400 ln (x - 20) - 200 + C

Solution steps

\int (2 x - 40) ln (x - 20) d x

Expand

(2 x - 40) ln (x - 20) : 2 x ln (x - 20) - 40 ln (x - 20)

= \int 2 x ln (x - 20) - 40 ln (x - 20) 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

= \int 2 x ln (x - 20) d x - \int 40 ln (x - 20) d x

\int 2 x ln (x - 20) d x = 2 (ln (x - 20) (\frac{1}{2} (x - 20)^{2} + 20 (x - 20)) - \frac{1}{4} (x^{2} + 40 x - 1200))

\int 40 ln (x - 20) d x = 40 (x ln (x - 20) - 20 ln (x - 20) - x + 20)

= 2 (ln (x - 20) (\frac{1}{2} (x - 20)^{2} + 20 (x - 20)) - \frac{1}{4} (x^{2} + 40 x - 1200)) - 40 (x ln (x - 20) - 20 ln (x - 20) - x + 20)

Simplify

2 (ln (x - 20) (\frac{1}{2} (x - 20)^{2} + 20 (x - 20)) - \frac{1}{4} (x^{2} + 40 x - 1200)) - 40 (x ln (x - 20) - 20 ln (x - 20) - x + 20) : x^{2} ln (x - 20) - \frac{1}{2} x^{2} + 20 x - 40 x ln (x - 20) + 400 ln (x - 20) - 200

= x^{2} ln (x - 20) - \frac{1}{2} x^{2} + 20 x - 40 x ln (x - 20) + 400 ln (x - 20) - 200

Add a constant to the solution

= x^{2} ln (x - 20) - \frac{1}{2} x^{2} + 20 x - 40 x ln (x - 20) + 400 ln (x - 20) - 200 + C

\int \frac{1}{( cos ( x ) - sin ( x ) ) ^{2}} d x

\int 3^{x} + \frac{1}{x} - x^{3} d x

\frac{d}{d x} (\frac{8 x}{x ^{2} + 3})

\frac{d}{d x} [f (x^{2})] = x^{5}

\frac{\partial}{\partial x} (x^{2} + 2 x y + 3 y^{4})

What is the integral of (2x-40)ln(x-20) ?
The integral of (2x-40)ln(x-20) is x^2ln(x-20)-1/2 x^2+20x-40xln(x-20)+400ln(x-20)-200+C