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

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

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

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

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

Decimal Notation

6.39228 \dots

Solution steps

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

Take the constant out:

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

= 2 \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}

= 2 [\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}

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

Compute the boundaries:

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

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

\int y x e^{x y} d y

e^{- 2 x}, 0

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

\int \frac{s \cdot ln ( s )}{( 1 - s ^{2} ) ^{\frac{1}{2}}} d s

\frac{d}{d x} (ln (x^{2} + y^{2} + z^{2}))

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