What is the integral from 1 to 2 of 8ln(x)-2x^2ln(x) ?

The integral from 1 to 2 of 8ln(x)-2x^2ln(x) is -58/9+32/3 ln(2)

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integral from 1 to 2 of 8ln(x)-2x^2ln(x)

\int_{1}^{2} 8 ln (x) - 2 x^{2} ln (x) d x

- \frac{58}{9} + \frac{32}{3} ln (2)

Decimal Notation

0.94912 \dots

Solution steps

\int_{1}^{2} 8 ln (x) - 2 x^{2} 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_{1}^{2} 8 ln (x) d x - \int_{1}^{2} 2 x^{2} ln (x) d x

\int_{1}^{2} 8 ln (x) d x = 8 (2 ln (2) - 1)

\int_{1}^{2} 2 x^{2} ln (x) d x = 2 (- \frac{7}{9} + \frac{8}{3} ln (2))

= 8 (2 ln (2) - 1) - 2 (- \frac{7}{9} + \frac{8}{3} ln (2))

Simplify

8 (2 ln (2) - 1) - 2 (- \frac{7}{9} + \frac{8}{3} ln (2)) : - \frac{58}{9} + \frac{32}{3} ln (2)

= - \frac{58}{9} + \frac{32}{3} ln (2)

\frac{\partial}{\partial y} (e^{x} ln (1 + y))

f (x) = x sin (x)

y^{^{''}} + \frac{1}{4} y^{^{'}} + 2 y = 2 cos (4123 t)

\int \frac{1 - 4 x ^{2}}{x} d x

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

What is the integral from 1 to 2 of 8ln(x)-2x^2ln(x) ?
The integral from 1 to 2 of 8ln(x)-2x^2ln(x) is -58/9+32/3 ln(2)