What is the integral from 2 to 5 of x^2ln(3x) ?

The integral from 2 to 5 of x^2ln(3x) is -13+(ln((5^{125}*3^{117})/(256)))/3

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integral from 2 to 5 of x^2ln(3x)

\int_{2}^{5} x^{2} ln (3 x) d x

- 13 + \frac{ln ( \frac{5 ^{125} \cdot 3 ^{117}}{256} )}{3}

Decimal

95.05739 \dots

Solution steps

\int_{2}^{5} x^{2} ln (3 x) d x

Apply Integration By Parts

: [\frac{1}{3} x^{3} ln (3 x) - \int \frac{x ^{2}}{3} d x]_{2}^{5}

= [\frac{1}{3} x^{3} ln (3 x) - \int \frac{x ^{2}}{3} d x]_{2}^{5}

\int \frac{x ^{2}}{3} d x = \frac{x ^{3}}{9}

= [\frac{1}{3} x^{3} ln (3 x) - \frac{x ^{3}}{9}]_{2}^{5}

Compute the boundaries:

- 13 + \frac{- 8 ln ( 6 ) + 125 ln ( 15 )}{3}

= - 13 + \frac{- 8 ln ( 6 ) + 125 ln ( 15 )}{3}

Simplify

= - 13 + \frac{ln ( \frac{5 ^{125} \cdot 3 ^{117}}{256} )}{3}

x \frac{d y}{d x} + y = sec (x y)

\int x^{7} ln (1 + x) d x

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

e ln (2 x^{4} + a x^{2} y^{2 + 2 y^{4}})

f (x) = 10 x

What is the integral from 2 to 5 of x^2ln(3x) ?
The integral from 2 to 5 of x^2ln(3x) is -13+(ln((5^{125}*3^{117})/(256)))/3