What is the integral from 1 to e^4 of x^2ln(x) ?

The integral from 1 to e^4 of x^2ln(x) is (11e^{12}+1)/9

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integral from 1 to e^4 of x^2ln(x)

\int_{1}^{e^{4}} x^{2} ln (x) d x

\frac{11 e ^{12} + 1}{9}

Decimal Notation

198922.63395 \dots

Solution steps

\int_{1}^{e^{4}} x^{2} ln (x) d x

Apply Integration By Parts

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

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

\int \frac{x ^{2}}{3} d x = \frac{x ^{3}}{9}

= [\frac{1}{3} (x^{3} ln (x) - 3 \cdot \frac{x ^{3}}{9})]_{1}^{e^{4}}

3 \cdot \frac{x ^{3}}{9} = \frac{x ^{3}}{3}

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

Compute the boundaries:

\frac{11 e ^{12}}{9} + \frac{1}{9}

= \frac{11 e ^{12}}{9} + \frac{1}{9}

Simplify

= \frac{11 e ^{12} + 1}{9}

y e^{2 x} d x = (4 + e^{2 x}) d y

x y^{2} y^{^{'}} = y^{3} - x^{3}, y (1) = 2

n = 1 \sum \infty n (\frac{4}{5})^{n}

\int - 9 x e^{- 3 x} d x

(x^{2} + y^{2}) d x + 3 x y d y = 0

What is the integral from 1 to e^4 of x^2ln(x) ?
The integral from 1 to e^4 of x^2ln(x) is (11e^{12}+1)/9