What is the integral from 1 to 3 of 9ln(x)-x^2ln(x) ?

The integral from 1 to 3 of 9ln(x)-x^2ln(x) is 18ln(3)-136/9

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integral from 1 to 3 of 9ln(x)-x^2ln(x)

\int_{1}^{3} 9 ln (x) - x^{2} ln (x) d x

18 ln (3) - \frac{136}{9}

Decimal

4.66391 \dots

Solution steps

\int_{1}^{3} 9 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}^{3} 9 ln (x) d x - \int_{1}^{3} x^{2} ln (x) d x

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

\int_{1}^{3} x^{2} ln (x) d x = 9 ln (3) - \frac{26}{9}

= 9 (3 ln (3) - 2) - (9 ln (3) - \frac{26}{9})

Simplify

9 (3 ln (3) - 2) - (9 ln (3) - \frac{26}{9}) : 18 ln (3) - \frac{136}{9}

= 18 ln (3) - \frac{136}{9}

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What is the integral from 1 to 3 of 9ln(x)-x^2ln(x) ?
The integral from 1 to 3 of 9ln(x)-x^2ln(x) is 18ln(3)-136/9