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

The integral from 1 to e^4 of x^5ln(x) is (23e^{24}+1)/(36)

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

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

\frac{23 e ^{24} + 1}{36}

Decimal Notation

16923605805.205528

Solution steps

\int_{1}^{e^{4}} 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^{4}}

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

\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^{4}}

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

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

Compute the boundaries:

\frac{23 e ^{24}}{36} + \frac{1}{36}

= \frac{23 e ^{24}}{36} + \frac{1}{36}

Simplify

= \frac{23 e ^{24} + 1}{36}

\frac{\partial}{\partial y} (y arctan (\frac{x}{z}))

h (x) = 4 x^{7}

\int_{2}^{5} \frac{3 x + 3}{x ^{2}} d x

x \to 1 lim (\frac{7 x ^{2} - 4 x - 3}{3 x ^{2} - 4 x + 1})

\frac{d}{d x} (94 - 10 cos (\frac{π}{12} (x - 2)))

What is the integral from 1 to e^4 of x^5ln(x) ?
The integral from 1 to e^4 of x^5ln(x) is (23e^{24}+1)/(36)