What is the integral from 1 to e of x^3(ln(x))^2 ?

The integral from 1 to e of x^3(ln(x))^2 is (5e^4-1)/(32)

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integral from 1 to e of x^3(ln(x))^2

\int_{1}^{e} x^{3} (ln (x))^{2} d x

\frac{5 e ^{4} - 1}{32}

Decimal Notation

8.49971 \dots

Solution steps

\int_{1}^{e} x^{3} (ln (x))^{2} d x

Simplify

= \int_{1}^{e} x^{3} ln^{2} (x) d x

Apply Integration By Parts

: [\frac{1}{4} (x^{4} ln^{2} (x) - 4 \cdot \int \frac{x ^{3} ln ( x )}{2} d x)]_{1}^{e}

= [\frac{1}{4} (x^{4} ln^{2} (x) - 4 \cdot \int \frac{x ^{3} ln ( x )}{2} d x)]_{1}^{e}

\int \frac{x ^{3} ln ( x )}{2} d x = \frac{1}{2} (\frac{1}{4} x^{4} ln (x) - \frac{x ^{4}}{16})

= [\frac{1}{4} (x^{4} ln^{2} (x) - 4 \cdot \frac{1}{2} (\frac{1}{4} x^{4} ln (x) - \frac{x ^{4}}{16}))]_{1}^{e}

4 \cdot \frac{1}{2} (\frac{1}{4} x^{4} ln (x) - \frac{x ^{4}}{16}) = 2 (\frac{1}{4} x^{4} ln (x) - \frac{x ^{4}}{16})

= [\frac{1}{4} (x^{4} ln^{2} (x) - 2 (\frac{1}{4} x^{4} ln (x) - \frac{x ^{4}}{16}))]_{1}^{e}

Compute the boundaries:

\frac{5 e ^{4}}{32} - \frac{1}{32}

= \frac{5 e ^{4}}{32} - \frac{1}{32}

Simplify

= \frac{5 e ^{4} - 1}{32}

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What is the integral from 1 to e of x^3(ln(x))^2 ?
The integral from 1 to e of x^3(ln(x))^2 is (5e^4-1)/(32)