What is the integral from 1 to 2 of 4ln(x)-x^2ln(x) ?

The integral from 1 to 2 of 4ln(x)-x^2ln(x) is -29/9+16/3 ln(2)

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integral from 1 to 2 of 4ln(x)-x^2ln(x)

\int_{1}^{2} 4 ln (x) - x^{2} ln (x) d x

- \frac{29}{9} + \frac{16}{3} ln (2)

Decimal Notation

0.47456 \dots

Solution steps

\int_{1}^{2} 4 ln (x) - 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} 4 ln (x) d x - \int_{1}^{2} x^{2} ln (x) d x

\int_{1}^{2} 4 ln (x) d x = 4 (2 ln (2) - 1)

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

= 4 (2 ln (2) - 1) - (- \frac{7}{9} + \frac{8}{3} ln (2))

Simplify

4 (2 ln (2) - 1) - (- \frac{7}{9} + \frac{8}{3} ln (2)) : - \frac{29}{9} + \frac{16}{3} ln (2)

= - \frac{29}{9} + \frac{16}{3} ln (2)

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What is the integral from 1 to 2 of 4ln(x)-x^2ln(x) ?
The integral from 1 to 2 of 4ln(x)-x^2ln(x) is -29/9+16/3 ln(2)