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

The integral from 0 to 1/2 of 1/(1+x^3) is 1/3 ln(3/2)+1/3 (1/2 ln(4/3)+pi/(2sqrt(3)))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\int_{0}^{\frac{1}{2}} \frac{1}{1 + x ^{3}} d x

\frac{1}{3} ln (\frac{3}{2}) + \frac{1}{3} (\frac{1}{2} ln (\frac{4}{3}) + \frac{π}{2 3})

Decimal

0.48540 \dots

Solution steps

\int_{0}^{\frac{1}{2}} \frac{1}{1 + x ^{3}} d x

Take the partial fraction of

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int_{0}^{\frac{1}{2}} \frac{- x + 2}{3 ( x ^{2} - x + 1 )} d x = \frac{1}{3} (\frac{1}{2} ln (\frac{4}{3}) + \frac{π}{2 3})

= \frac{1}{3} ln (\frac{3}{2}) + \frac{1}{3} (\frac{1}{2} ln (\frac{4}{3}) + \frac{π}{2 3})

\int_{0}^{1} 3 r^{2} e^{r^{3}} d r

\int_{2}^{5} 2 x^{2} d x

\int_{- 1}^{1} \frac{x}{5 - 4 x} d x

\int_{1}^{3} \frac{x - 1}{x} d x

\int_{1}^{\infty} \frac{2}{3 x + 9} d x

What is the integral from 0 to 1/2 of 1/(1+x^3) ?
The integral from 0 to 1/2 of 1/(1+x^3) is 1/3 ln(3/2)+1/3 (1/2 ln(4/3)+pi/(2sqrt(3)))