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

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

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

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

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

Decimal

3.19614 \dots

Solution steps

\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}

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

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

Compute the boundaries:

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

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

\frac{d}{d x} (6 x^{- 1})

d er i v a t i v e e^{(c o s (t) + l n (t))}

x \to \infty lim (\frac{1}{2 x})

\frac{d}{d x} (ln (ln (2 x^{3})))

d er i v a t i v e \frac{x - 6}{x + 6}

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