What is the integral from 1 to e^6 of x^4ln(x) ?

The integral from 1 to e^6 of x^4ln(x) is (29e^{30}+1)/(25)

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integral from 1 to e^6 of x^4ln(x)

\int_{1}^{e^{6}} x^{4} ln (x) d x

\frac{29 e ^{30} + 1}{25}

Decimal Notation

1.23963 E 13

Solution steps

\int_{1}^{e^{6}} 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^{6}}

= [\frac{1}{5} (x^{5} ln (x) - 5 \cdot \int \frac{x ^{4}}{5} d x)]_{1}^{e^{6}}

\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^{6}}

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

= [\frac{1}{5} (x^{5} ln (x) - \frac{x ^{5}}{5})]_{1}^{e^{6}}

Compute the boundaries:

\frac{29 e ^{30}}{25} + \frac{1}{25}

= \frac{29 e ^{30}}{25} + \frac{1}{25}

Simplify

= \frac{29 e ^{30} + 1}{25}

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What is the integral from 1 to e^6 of x^4ln(x) ?
The integral from 1 to e^6 of x^4ln(x) is (29e^{30}+1)/(25)