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+6x^{1/2)}

解

\int 1 + 6 x^{\frac{1}{2}} d x

解

\frac{2}{135} (9 x - 1) (6 x + 1)^{\frac{3}{2}} + C

解答ステップ

\int 1 + 6 x^{\frac{1}{2}} d x

u置換積分法を適用する

= \int 1 + 6 u \cdot 2 u d u

定数を除く:

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

= 2 \cdot \int 1 + 6 u u d u

u置換積分法を適用する

= 2 \cdot \int \frac{v ( v - 1 )}{36} d v

定数を除く:

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

= 2 \cdot \frac{1}{36} \cdot \int v (v - 1) d v

拡張

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

= 2 \cdot \frac{1}{36} \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

= 2 \cdot \frac{1}{36} (\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}}

= 2 \cdot \frac{1}{36} (\frac{2}{5} v^{\frac{5}{2}} - \frac{2}{3} v^{\frac{3}{2}})

代用を戻す

= 2 \cdot \frac{1}{36} (\frac{2}{5} (1 + 6 x^{\frac{1}{2}})^{\frac{5}{2}} - \frac{2}{3} (1 + 6 x^{\frac{1}{2}})^{\frac{3}{2}})

簡素化

2 \cdot \frac{1}{36} (\frac{2}{5} (1 + 6 x^{\frac{1}{2}})^{\frac{5}{2}} - \frac{2}{3} (1 + 6 x^{\frac{1}{2}})^{\frac{3}{2}}) : \frac{2}{135} (9 x - 1) (6 x + 1)^{\frac{3}{2}}

= \frac{2}{135} (9 x - 1) (6 x + 1)^{\frac{3}{2}}

定数を解答に追加する

= \frac{2}{135} (9 x - 1) (6 x + 1)^{\frac{3}{2}} + C

グラフ

人気の例

derivative f(x)=x*sin(x)

d er i v a t i v e f (x) = x \cdot sin (x)

f^{^{'}} (x) = x^{2} - x

inverselaplace 2+(4(s-4))/(s^2+16)

in v erse l a pl a ce 2 + \frac{4 ( s - 4 )}{s ^{2} + 16}

derivative of-3sin(2x)

\frac{d}{d x} (- 3 sin (2 x))

sum from n=0 to infinity of 1/(2^{2^n)}

n = 0 \sum \infty \frac{1}{2 ^{2^{n}}}