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

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

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integral from 1 to e of 4x^3ln(x)

\int_{1}^{e} 4 x^{3} ln (x) d x

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

Decimal Notation

41.19861 \dots

Solution steps

\int_{1}^{e} 4 x^{3} ln (x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \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}

= 4 [\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}

= 4 [\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}

= 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}

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

Simplify

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

- \frac{1}{2} x

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\int \frac{3}{2} e^{- x} cos (2 x) d x

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