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 sqrt(1+\sqrt{2x)}

解

\int 1 + 2 x d x

解

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

解答ステップ

\int 1 + 2 x d x

u置換積分法を適用する

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

定数を除く:

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

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

u置換積分法を適用する

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

定数を除く:

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

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

拡張

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

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

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

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

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

= \frac{1}{2} \cdot 2 (\frac{2}{5} v^{\frac{5}{2}} - \frac{2}{3} v^{\frac{3}{2}})

代用を戻す

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

簡素化

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

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

定数を解答に追加する

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

グラフ

人気の例

integral of (e^{2y}-ycos(x)y)

\int (e^{2 y} - y cos (x) y) d x

sin(y)csc(x)((dy)/(dx))=1

sin (y) csc (x) (\frac{d y}{d x}) = 1

tangent f(x)=x^2-8x+9,\at x=3

t an g e n t f (x) = x^{2} - 8 x + 9, at x = 3

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

\int_{0}^{1} \frac{2}{1 + t ^{2}} d t

integral of (12)/(1-cos(4x))

\int \frac{12}{1 - cos ( 4 x )} d x