What is the integral from 1 to e^2 of x^5ln(x) ?

The integral from 1 to e^2 of x^5ln(x) is (11e^{12}+1)/(36)

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integral from 1 to e^2 of x^5ln(x)

\int_{1}^{e^{2}} x^{5} ln (x) d x

\frac{11 e ^{12} + 1}{36}

Decimal Notation

49730.65848 \dots

Solution steps

\int_{1}^{e^{2}} 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^{2}}

= [\frac{1}{6} (x^{6} ln (x) - 6 \cdot \int \frac{x ^{5}}{6} d x)]_{1}^{e^{2}}

\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^{2}}

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

= [\frac{1}{6} (x^{6} ln (x) - \frac{x ^{6}}{6})]_{1}^{e^{2}}

Compute the boundaries:

\frac{11 e ^{12}}{36} + \frac{1}{36}

= \frac{11 e ^{12}}{36} + \frac{1}{36}

Simplify

= \frac{11 e ^{12} + 1}{36}

\int x y e^{x y^{2}} d y

\frac{d}{d x} (\frac{6 x}{x ^{2} - 4})

\int_{0}^{N} e^{- s t} e^{- 4 t} d t

x \to + 0 + lim (\frac{5 x}{sin ( x )})

\frac{1}{x} \cdot 3 e^{x}

What is the integral from 1 to e^2 of x^5ln(x) ?
The integral from 1 to e^2 of x^5ln(x) is (11e^{12}+1)/(36)