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

The integral from 1 to e^2 of x^2ln(x) is (5e^6+1)/9

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{5 e ^{6} + 1}{9}

Decimal

224.23821 \dots

Solution steps

\int_{1}^{e^{2}} 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^{2}}

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

\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^{2}}

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

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

Compute the boundaries:

\frac{5 e ^{6}}{9} + \frac{1}{9}

= \frac{5 e ^{6}}{9} + \frac{1}{9}

Simplify

= \frac{5 e ^{6} + 1}{9}

\frac{x + 4}{x ^{3} + x - 5}

y = 2 x, x = 5, y = \frac{2}{x}

\int \frac{x + 1}{x ^{3} + x ^{2} - 6 x} d x

y = x^{2}, at x = - 0.5

\frac{d}{d x} (1 + x^{2} y^{2})

What is the integral from 1 to e^2 of x^2ln(x) ?
The integral from 1 to e^2 of x^2ln(x) is (5e^6+1)/9