What is the integral of (x^2-10x+24)ln(x) ?

The integral of (x^2-10x+24)ln(x) is 1/3 x^3ln(x)-(x^3)/9-10(1/2 x^2ln(x)-(x^2)/4)+24(xln(x)-x)+C

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integral of (x^2-10x+24)ln(x)

\int (x^{2} - 10 x + 24) ln (x) d x

\frac{1}{3} x^{3} ln (x) - \frac{x ^{3}}{9} - 10 (\frac{1}{2} x^{2} ln (x) - \frac{x ^{2}}{4}) + 24 (x ln (x) - x) + C

Solution steps

\int (x^{2} - 10 x + 24) ln (x) d x

Expand

(x^{2} - 10 x + 24) ln (x) : x^{2} ln (x) - 10 x ln (x) + 24 ln (x)

= \int x^{2} ln (x) - 10 x ln (x) + 24 ln (x) 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 x^{2} ln (x) d x - \int 10 x ln (x) d x + \int 24 ln (x) d x

\int x^{2} ln (x) d x = \frac{1}{3} x^{3} ln (x) - \frac{x ^{3}}{9}

\int 10 x ln (x) d x = 10 (\frac{1}{2} x^{2} ln (x) - \frac{x ^{2}}{4})

\int 24 ln (x) d x = 24 (x ln (x) - x)

= \frac{1}{3} x^{3} ln (x) - \frac{x ^{3}}{9} - 10 (\frac{1}{2} x^{2} ln (x) - \frac{x ^{2}}{4}) + 24 (x ln (x) - x)

Add a constant to the solution

= \frac{1}{3} x^{3} ln (x) - \frac{x ^{3}}{9} - 10 (\frac{1}{2} x^{2} ln (x) - \frac{x ^{2}}{4}) + 24 (x ln (x) - x) + C

f (x) = \frac{1}{2 + 3 x}, (1, \frac{1}{5})

e^{- 2 t} t

e^{x}, - 4

\frac{\partial}{\partial y} ((2 x + 3 y)^{10})

\frac{1 - 3 t}{4 + t}

What is the integral of (x^2-10x+24)ln(x) ?
The integral of (x^2-10x+24)ln(x) is 1/3 x^3ln(x)-(x^3)/9-10(1/2 x^2ln(x)-(x^2)/4)+24(xln(x)-x)+C