What is the integral from 1 to e^3 of x^3ln(x) ?

The integral from 1 to e^3 of x^3ln(x) is (11e^{12}+1)/(16)

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integral from 1 to e^3 of x^3ln(x)

\int_{1}^{e^{3}} x^{3} ln (x) d x

\frac{11 e ^{12} + 1}{16}

Decimal

111893.98160 \dots

Solution steps

\int_{1}^{e^{3}} x^{3} ln (x) d x

Apply Integration By Parts

: [\frac{1}{4} (x^{4} ln (x) - 4 \cdot \int \frac{x ^{3}}{4} d x)]_{1}^{e^{3}}

= [\frac{1}{4} (x^{4} ln (x) - 4 \cdot \int \frac{x ^{3}}{4} d x)]_{1}^{e^{3}}

\int \frac{x ^{3}}{4} d x = \frac{x ^{4}}{16}

= [\frac{1}{4} (x^{4} ln (x) - 4 \cdot \frac{x ^{4}}{16})]_{1}^{e^{3}}

4 \cdot \frac{x ^{4}}{16} = \frac{x ^{4}}{4}

= [\frac{1}{4} (x^{4} ln (x) - \frac{x ^{4}}{4})]_{1}^{e^{3}}

Compute the boundaries:

\frac{11 e ^{12}}{16} + \frac{1}{16}

= \frac{11 e ^{12}}{16} + \frac{1}{16}

Simplify

= \frac{11 e ^{12} + 1}{16}

y^{^{''}} - 2 y^{^{'}} - 3 y = 4 x - 5 + 6 x e^{2 x}

\int 3 cos (x) sin (x) d x

n = 1 \sum \infty \frac{n ^{2}}{2 ^{n}}

\frac{d}{d x} ((6 x - 1)^{2})

\frac{\partial}{\partial x} ((2 x y) (x^{2} + y^{2}))

What is the integral from 1 to e^3 of x^3ln(x) ?
The integral from 1 to e^3 of x^3ln(x) is (11e^{12}+1)/(16)