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 4t^3sqrt(t^2+1)

解

\int 4 t^{3} t^{2} + 1 d t

解

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

解答ステップ

\int 4 t^{3} t^{2} + 1 d t

定数を除く:

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

= 4 \cdot \int t^{3} t^{2} + 1 d t

u置換積分法を適用する

= 4 \cdot \int \frac{( u - 1 ) u}{2} d u

定数を除く:

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

= 4 \cdot \frac{1}{2} \cdot \int (u - 1) u d u

拡張

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

= 4 \cdot \frac{1}{2} \cdot \int u^{\frac{3}{2}} - u 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

= 4 \cdot \frac{1}{2} (\int u^{\frac{3}{2}} d u - \int u d u)

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

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

= 4 \cdot \frac{1}{2} (\frac{2}{5} u^{\frac{5}{2}} - \frac{2}{3} u^{\frac{3}{2}})

代用を戻す

u = t^{2} + 1

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

簡素化

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

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

定数を解答に追加する

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

グラフ

人気の例

y^'=(2y)/x-x^2y^2

y^{^{'}} = \frac{2 y}{x} - x^{2} y^{2}

limit as x approaches infinity of x^{sin(x)}

x \to \infty lim (x^{s i n (x)})

テイラー cos(x)+sin(x)

t a y l or cos (x) + sin (x)

(\partial)/(\partial x)(-2ycos(x))

\frac{\partial}{\partial x} (- 2 y cos (x))

derivative tan(9x)

d er i v a t i v e tan (9 x)