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 1/(sqrt(1+\sqrt[3]{x))}

解

\int \frac{1}{1 + 3 x} d x

解

\frac{6}{5} (1 + 3 x)^{\frac{5}{2}} - 4 (1 + 3 x)^{\frac{3}{2}} + 6 1 + 3 x + C

解答ステップ

\int \frac{1}{1 + 3 x} d x

u置換積分法を適用する

= \int \frac{3 u ^{2}}{1 + u} d u

定数を除く:

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

= 3 \cdot \int \frac{u ^{2}}{1 + u} d u

u置換積分法を適用する

= 3 \cdot \int \frac{( v - 1 ) ^{2}}{v} d v

拡張

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

= 3 \cdot \int v^{\frac{3}{2}} - 2 v + \frac{1}{v} d v

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

= 3 (\int v^{\frac{3}{2}} d v - \int 2 v d v + \int \frac{1}{v} d v)

\int v^{\frac{3}{2}} d v = \frac{2}{5} v^{\frac{5}{2}}

\int 2 v d v = \frac{4}{3} v^{\frac{3}{2}}

\int \frac{1}{v} d v = 2 v

= 3 (\frac{2}{5} v^{\frac{5}{2}} - \frac{4}{3} v^{\frac{3}{2}} + 2 v)

代用を戻す

= 3 (\frac{2}{5} (1 + 3 x)^{\frac{5}{2}} - \frac{4}{3} (1 + 3 x)^{\frac{3}{2}} + 2 1 + 3 x)

簡素化

3 (\frac{2}{5} (1 + 3 x)^{\frac{5}{2}} - \frac{4}{3} (1 + 3 x)^{\frac{3}{2}} + 2 1 + 3 x) : \frac{6}{5} (1 + 3 x)^{\frac{5}{2}} - 4 (1 + 3 x)^{\frac{3}{2}} + 6 1 + 3 x

= \frac{6}{5} (1 + 3 x)^{\frac{5}{2}} - 4 (1 + 3 x)^{\frac{3}{2}} + 6 1 + 3 x

定数を解答に追加する

= \frac{6}{5} (1 + 3 x)^{\frac{5}{2}} - 4 (1 + 3 x)^{\frac{3}{2}} + 6 1 + 3 x + C

グラフ

人気の例

limit as x approaches 1 of 7/(x^3-1)

x \to 1 lim (\frac{7}{x ^{3} - 1})

limit as x approaches 0 of 1/x-1/(x^2-x)

x \to 0 lim (\frac{1}{x} - \frac{1}{x ^{2} - x})

(\partial)/(\partial z)(5xsin(y-z))

\frac{\partial}{\partial z} (5 x sin (y - z))

面積 f(x)= 1/(7(1+x^2)),g(x)= 1/(14x^2)

a re a f (x) = \frac{1}{7 ( 1 + x ^{2} )}, g (x) = \frac{1}{14 x ^{2}}

integral of (x^2+2)/(x^3+3x^2+2x)

\int \frac{x ^{2} + 2}{x ^{3} + 3 x ^{2} + 2 x} d x