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

The integral from 1 to e^2 of x^4ln(x) is (9e^{10}+1)/(25)

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

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

\frac{9 e ^{10} + 1}{25}

Decimal

7929.56768 \dots

Solution steps

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

Apply Integration By Parts

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

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

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

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

5 \cdot \frac{x ^{5}}{25} = \frac{x ^{5}}{5}

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

Compute the boundaries:

\frac{9 e ^{10}}{25} + \frac{1}{25}

= \frac{9 e ^{10}}{25} + \frac{1}{25}

Simplify

= \frac{9 e ^{10} + 1}{25}

\int 5 x e^{- 3 x} d x

\frac{d y}{d x} = x + 9

x \to 2 - lim (\frac{x}{x ^{2} - 4})

f (x) = \frac{1}{x ^{2} - 9}

y = (tan (- 1 (9 x))) 2

What is the integral from 1 to e^2 of x^4ln(x) ?
The integral from 1 to e^2 of x^4ln(x) is (9e^{10}+1)/(25)