ja

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

人気のある微分積分 >

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

解

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

解

- 13 + \frac{96 ln ( 2 ) + 144}{16}

+1

十進法表記

0.15888 \dots

解答ステップ

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

拡張

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

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

拡張

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

総和規則を適用する:

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

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

\int_{0}^{1} \frac{1}{1 + 3 x + 3 x ^{2} + x ^{3}} d x = \frac{3}{8}

\int_{0}^{1} \frac{3 x}{1 + 3 x + 3 x ^{2} + x ^{3}} d x = \frac{3}{8}

\int_{0}^{1} \frac{3 x ^{2}}{1 + 3 x + 3 x ^{2} + x ^{3}} d x = 3 (\frac{16 ln ( 2 ) + 30}{16} - \frac{5}{2})

\int_{0}^{1} \frac{x ^{3}}{1 + 3 x + 3 x ^{2} + x ^{3}} d x = \frac{- 48 ln ( 2 ) - 54}{16} + \frac{11}{2}

= \frac{3}{8} - \frac{3}{8} + 3 (\frac{16 ln ( 2 ) + 30}{16} - \frac{5}{2}) - (\frac{- 48 ln ( 2 ) - 54}{16} + \frac{11}{2})

簡素化

\frac{3}{8} - \frac{3}{8} + 3 (\frac{16 ln ( 2 ) + 30}{16} - \frac{5}{2}) - (\frac{- 48 ln ( 2 ) - 54}{16} + \frac{11}{2}) : - 13 + \frac{96 ln ( 2 ) + 144}{16}

= - 13 + \frac{96 ln ( 2 ) + 144}{16}

グラフ

人気の例

integral from 1 to 2 of (5x^4-4x^3)

\int_{1}^{2} (5 x^{4} - 4 x^{3}) d x

integral from 2 to sin(x^2) of 1/(1+t^2)

\int_{2}^{s i n (x^{2})} \frac{1}{1 + t ^{2}} d t

integral from-pi/4 to pi/4 of sec^2(t)

\int_{- \frac{π}{4}}^{\frac{π}{4}} sec^{2} (t) d t

integral from-1 to 1 of xy^2

\int_{- 1}^{1} x y^{2} d y

integral from 0 to 1 of sin(pi)t

\int_{0}^{1} sin (π) t d t