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

The integral from 1 to e^5 of x^5ln(x) is (29e^{30}+1)/(36)

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

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

\frac{29 e ^{30} + 1}{36}

Decimal

8608548968450.277

Solution steps

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

Apply Integration By Parts

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

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

\int \frac{x ^{5}}{6} d x = \frac{x ^{6}}{36}

= [\frac{1}{6} (x^{6} ln (x) - 6 \cdot \frac{x ^{6}}{36})]_{1}^{e^{5}}

6 \cdot \frac{x ^{6}}{36} = \frac{x ^{6}}{6}

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

Compute the boundaries:

\frac{29 e ^{30}}{36} + \frac{1}{36}

= \frac{29 e ^{30}}{36} + \frac{1}{36}

Simplify

= \frac{29 e ^{30} + 1}{36}

\frac{\partial}{\partial x} (80 x^{\frac{1}{2}} z^{\frac{1}{2}})

y = x^{2}, y = 2 x + 3

\int_{0}^{8} \frac{7}{3} x d x

x \to 0 lim (\frac{2}{3 cos ( x )})

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

What is the integral from 1 to e^5 of x^5ln(x) ?
The integral from 1 to e^5 of x^5ln(x) is (29e^{30}+1)/(36)