What is the integral from 1 to 2 of 6x^4ln(x) ?

The integral from 1 to 2 of 6x^4ln(x) is 6(-31/25+32/5 ln(2))

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integral from 1 to 2 of 6x^4ln(x)

\int_{1}^{2} 6 x^{4} ln (x) d x

6 (- \frac{31}{25} + \frac{32}{5} ln (2))

Decimal Notation

19.17685 \dots

Solution steps

\int_{1}^{2} 6 x^{4} ln (x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int_{1}^{2} x^{4} ln (x) d x

Apply Integration By Parts

: [\frac{1}{5} x^{5} ln (x) - \int \frac{x ^{4}}{5} d x]_{1}^{2}

= 6 [\frac{1}{5} x^{5} ln (x) - \int \frac{x ^{4}}{5} d x]_{1}^{2}

\int \frac{x ^{4}}{5} d x = \frac{x ^{5}}{25}

= 6 [\frac{1}{5} x^{5} ln (x) - \frac{x ^{5}}{25}]_{1}^{2}

Compute the boundaries:

- \frac{31}{25} + \frac{32}{5} ln (2)

= 6 (- \frac{31}{25} + \frac{32}{5} ln (2))

f (x) = x^{3} ln (x)

\frac{d}{d x} ((tan (x))^{e^{x}})

\int x^{2} 5 x + 6 d x

x \to 0 lim (C e^{- 4 x})

x \to - 1 lim (\frac{x ^{2} + 1}{x})

What is the integral from 1 to 2 of 6x^4ln(x) ?
The integral from 1 to 2 of 6x^4ln(x) is 6(-31/25+32/5 ln(2))