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 of x\sqrt[5]{(x+1)^2}

解

\int x 5 (x + 1)^{2} d x

解

\frac{5}{12} (x + 1)^{\frac{12}{5}} - \frac{5}{7} (x + 1)^{\frac{7}{5}} + C

解答ステップ

\int x 5 (x + 1)^{2} d x

5 (x + 1)^{2} = (x + 1)^{\frac{2}{5}}, （ (x + 1) \geq 0

と仮定

）

= \int x (x + 1)^{\frac{2}{5}} d x

u置換積分法を適用する

= \int (u - 1) u^{\frac{2}{5}} d u

拡張

(u - 1) u^{\frac{2}{5}} : u^{\frac{7}{5}} - u^{\frac{2}{5}}

= \int u^{\frac{7}{5}} - u^{\frac{2}{5}} d u

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 u^{\frac{7}{5}} d u - \int u^{\frac{2}{5}} d u

\int u^{\frac{7}{5}} d u = \frac{5}{12} u^{\frac{12}{5}}

\int u^{\frac{2}{5}} d u = \frac{5}{7} u^{\frac{7}{5}}

= \frac{5}{12} u^{\frac{12}{5}} - \frac{5}{7} u^{\frac{7}{5}}

代用を戻す

u = x + 1

= \frac{5}{12} (x + 1)^{\frac{12}{5}} - \frac{5}{7} (x + 1)^{\frac{7}{5}}

定数を解答に追加する

= \frac{5}{12} (x + 1)^{\frac{12}{5}} - \frac{5}{7} (x + 1)^{\frac{7}{5}} + C

グラフ

人気の例

(x^2-25ln(x^2))^'

(x^{2} - 25 ln (x^{2}))^{^{'}}

integral of (10)/(15sin(x)-8cos(x))

\int \frac{10}{15 sin ( x ) - 8 cos ( x )} d x

integral of 6/(xsqrt(36-(ln(x))^2))

\int \frac{6}{x 36 - ( ln ( x ) ) ^{2}} d x

tangent f(x)=6xcos(x),(pi,-6pi)

t an g e n t f (x) = 6 x cos (x), (π, - 6 π)

derivative of e^{2x}-e^{-2x}-4x

\frac{d}{d x} (e^{2 x} - e^{- 2 x} - 4 x)