What is the integral from 1 to 2 of x^{1/8}ln(x) ?

The integral from 1 to 2 of x^{1/8}ln(x) is (16*2^{1/8})/9 ln(2)+(-128*2^{1/8}+64)/(81)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 1 to 2 of x^{1/8}ln(x)

\int_{1}^{2} x^{\frac{1}{8}} ln (x) d x

\frac{16 \cdot 2 ^{\frac{1}{8}}}{9} ln (2) + \frac{- 128 \cdot 2 ^{\frac{1}{8}} + 64}{81}

Decimal

0.41064 \dots

Solution steps

\int_{1}^{2} x^{\frac{1}{8}} ln (x) d x

Apply Integration By Parts

= [\frac{8}{9} x^{\frac{9}{8}} ln (x) - \int \frac{8}{9} x^{\frac{1}{8}} d x]_{1}^{2}

\int \frac{8}{9} x^{\frac{1}{8}} d x = \frac{64}{81} x^{\frac{9}{8}}

= [\frac{8}{9} x^{\frac{9}{8}} ln (x) - \frac{64}{81} x^{\frac{9}{8}}]_{1}^{2}

Compute the boundaries:

\frac{16 \cdot 2 ^{\frac{1}{8}}}{9} ln (2) - \frac{128 \cdot 2 ^{\frac{1}{8}}}{81} + \frac{64}{81}

= \frac{16 \cdot 2 ^{\frac{1}{8}}}{9} ln (2) - \frac{128 \cdot 2 ^{\frac{1}{8}}}{81} + \frac{64}{81}

Simplify

= \frac{16 \cdot 2 ^{\frac{1}{8}}}{9} ln (2) + \frac{- 128 \cdot 2 ^{\frac{1}{8}} + 64}{81}

What is the integral from 1 to 2 of x^{1/8}ln(x) ?
The integral from 1 to 2 of x^{1/8}ln(x) is (16*2^{1/8})/9 ln(2)+(-128*2^{1/8}+64)/(81)