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

The integral from 1 to 2 of x^4(ln(x))^2 is 62/125+32/5 ln^2(2)-64/25 ln(2)

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

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

\frac{62}{125} + \frac{32}{5} ln^{2} (2) - \frac{64}{25} ln (2)

Decimal Notation

1.79644 \dots

Solution steps

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

Apply Integration By Parts

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

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

\int \frac{2}{5} x^{4} ln (x) d x = \frac{2}{5} (\frac{1}{5} x^{5} ln (x) - \frac{x ^{5}}{25})

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

Compute the boundaries:

\frac{62}{125} + \frac{32}{5} ln^{2} (2) - \frac{64}{25} ln (2)

= \frac{62}{125} + \frac{32}{5} ln^{2} (2) - \frac{64}{25} ln (2)

\frac{d}{d x} (- \frac{1}{4 ( x + 1 ) ^{\frac{3}{2}}})

\int_{1}^{2} (\frac{3}{x}) d x

\frac{d}{d x} (\frac{- 5}{x})

\int \frac{1}{x ^{- 2}} d x

n = 6 \sum \infty e^{7 - 6 n}

What is the integral from 1 to 2 of x^4(ln(x))^2 ?
The integral from 1 to 2 of x^4(ln(x))^2 is 62/125+32/5 ln^2(2)-64/25 ln(2)