What is the integral from 1 to 3 of x^4(ln(x))^2 ?

The integral from 1 to 3 of x^4(ln(x))^2 is 484/125+243/5 ln^2(3)-486/25 ln(3)

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integral from 1 to 3 of x^4(ln(x))^2

\int_{1}^{3} x^{4} (ln (x))^{2} d x

\frac{484}{125} + \frac{243}{5} ln^{2} (3) - \frac{486}{25} ln (3)

Decimal

41.17269 \dots

Solution steps

\int_{1}^{3} 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}^{3}

= [\frac{1}{5} x^{5} ln^{2} (x) - \int \frac{2}{5} x^{4} ln (x) d x]_{1}^{3}

\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}^{3}

Compute the boundaries:

\frac{484}{125} + \frac{243}{5} ln^{2} (3) - \frac{486}{25} ln (3)

= \frac{484}{125} + \frac{243}{5} ln^{2} (3) - \frac{486}{25} ln (3)

8 e^{- 2 x}

f (x) = x - 2 sin (x)

\int x^{2} + 6 x + 45 d x

x \to 0 lim (\frac{2}{x ^{6}})

n = 1 \sum \infty \frac{4 ^{n}}{n ^{2}}

What is the integral from 1 to 3 of x^4(ln(x))^2 ?
The integral from 1 to 3 of x^4(ln(x))^2 is 484/125+243/5 ln^2(3)-486/25 ln(3)