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

The integral from 1 to e of x^3ln(x) is (3e^4+1)/(16)

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

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

\frac{3 e ^{4} + 1}{16}

Decimal Notation

10.29965 \dots

Solution steps

\int_{1}^{e} 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}

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

\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}

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

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

Compute the boundaries:

\frac{3 e ^{4}}{16} + \frac{1}{16}

= \frac{3 e ^{4}}{16} + \frac{1}{16}

Simplify

= \frac{3 e ^{4} + 1}{16}

\frac{d}{d x} (\frac{x}{20})

y = 7 e^{x} + x, (0, 7)

4

\int_{0}^{6} x^{2} - 6 x d x

x^{2}, x + 5

What is the integral from 1 to e of x^3ln(x) ?
The integral from 1 to e of x^3ln(x) is (3e^4+1)/(16)