What is the integral from 1 to e^5 of x^4ln(x) ?

The integral from 1 to e^5 of x^4ln(x) is (24e^{25}+1)/(25)

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integral from 1 to e^5 of x^4ln(x)

\int_{1}^{e^{5}} x^{4} ln (x) d x

\frac{24 e ^{25} + 1}{25}

Decimal Notation

69124703363.93034

Solution steps

\int_{1}^{e^{5}} 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^{5}}

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

\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^{5}}

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

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

Compute the boundaries:

\frac{24 e ^{25}}{25} + \frac{1}{25}

= \frac{24 e ^{25}}{25} + \frac{1}{25}

Simplify

= \frac{24 e ^{25} + 1}{25}

y = \frac{x ^{2} - 1}{x ^{2} + 1}

h (t) = 65 - 55 cos (\frac{π t}{15})

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

\int \frac{1}{x ^{3} x ^{2} - 49} d x

\frac{x ^{2} + 1}{x ^{2} - x - 6}

What is the integral from 1 to e^5 of x^4ln(x) ?
The integral from 1 to e^5 of x^4ln(x) is (24e^{25}+1)/(25)