What is the integral from 0 to 1 of x^2(ln(1/x))^3 ?

The integral from 0 to 1 of x^2(ln(1/x))^3 is 2/27

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1 of x^2(ln(1/x))^3

\int_{0}^{1} x^{2} (ln (\frac{1}{x}))^{3} d x

\frac{2}{27}

Decimal Notation

0.07407 \dots

Solution steps

\int_{0}^{1} x^{2} (ln (\frac{1}{x}))^{3} d x

Simplify

x^{2} (ln (\frac{1}{x}))^{3} : - x^{2} ln^{3} (x)

= \int_{0}^{1} - x^{2} ln^{3} (x) d x

Take the constant out:

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

= - \int_{0}^{1} x^{2} ln^{3} (x) d x

Apply Integration By Parts

: [\frac{1}{3} x^{3} ln^{3} (x) - \int x^{2} ln^{2} (x) d x]_{0}^{1}

= - [\frac{1}{3} x^{3} ln^{3} (x) - \int x^{2} ln^{2} (x) d x]_{0}^{1}

\int x^{2} ln^{2} (x) d x = \frac{1}{3} x^{3} ln^{2} (x) - \frac{2}{3} (\frac{1}{3} x^{3} ln (x) - \frac{x ^{3}}{9})

= - [\frac{1}{3} x^{3} ln^{3} (x) - (\frac{1}{3} x^{3} ln^{2} (x) - \frac{2}{3} (\frac{1}{3} x^{3} ln (x) - \frac{x ^{3}}{9}))]_{0}^{1}

Expand

\frac{1}{3} x^{3} ln^{3} (x) - (\frac{1}{3} x^{3} ln^{2} (x) - \frac{2}{3} (\frac{1}{3} x^{3} ln (x) - \frac{x ^{3}}{9})) : \frac{1}{3} x^{3} ln^{3} (x) - \frac{1}{3} x^{3} ln^{2} (x) + \frac{2}{9} x^{3} ln (x) - \frac{2 x ^{3}}{27}

= - [\frac{1}{3} x^{3} ln^{3} (x) - \frac{1}{3} x^{3} ln^{2} (x) + \frac{2}{9} x^{3} ln (x) - \frac{2 x ^{3}}{27}]_{0}^{1}

Compute the boundaries:

- \frac{2}{27}

= - (- \frac{2}{27})

Simplify

= \frac{2}{27}

\int_{0}^{2} 4 - x d x

\int_{0}^{2} - g \cdot cos (n \cdot π \cdot g) d g

\int_{- 1}^{2} y^{2} + 4 d y

\int_{0}^{2} 20 x d x

\int_{0}^{x^{2}} x^{2} e^{x^{2}} d x

What is the integral from 0 to 1 of x^2(ln(1/x))^3 ?
The integral from 0 to 1 of x^2(ln(1/x))^3 is 2/27