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

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

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

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

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

Decimal Notation

41.19861 \dots

Solution steps

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

Take the constant out:

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

= 4 \cdot \int_{1}^{e} t^{3} ln (t) d t

Apply Integration By Parts

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

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

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

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

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

= 4 [\frac{1}{4} (t^{4} ln (t) - \frac{t ^{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}

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