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^2)/((144+x^2)^2)

解

\int \frac{x ^{2}}{( 144 + x ^{2} ) ^{2}} d x

解

\frac{1}{12} (\frac{1}{2} arctan (\frac{x}{12}) - \frac{6 x}{144 + x ^{2}}) + C

解答ステップ

\int \frac{x ^{2}}{( 144 + x ^{2} ) ^{2}} d x

置換積分法を適用する

= \int \frac{u ^{2}}{12 ( u ^{2} + 1 ) ^{2}} d u

定数を除く:

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

= \frac{1}{12} \cdot \int \frac{u ^{2}}{( u ^{2} + 1 ) ^{2}} d u

以下の部分分数を得る:

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

= \frac{1}{12} \cdot \int \frac{1}{u ^{2} + 1} - \frac{1}{( u ^{2} + 1 ) ^{2}} 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

= \frac{1}{12} (\int \frac{1}{u ^{2} + 1} d u - \int \frac{1}{( u ^{2} + 1 ) ^{2}} d u)

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

\int \frac{1}{( u ^{2} + 1 ) ^{2}} d u = \frac{1}{2} (arctan (u) + \frac{u}{1 + u ^{2}})

= \frac{1}{12} (arctan (u) - \frac{1}{2} (arctan (u) + \frac{u}{1 + u ^{2}}))

代用を戻す

u = \frac{x}{12}

= \frac{1}{12} (arctan (\frac{x}{12}) - \frac{1}{2} (arctan (\frac{x}{12}) + \frac{\frac{x}{12}}{1 + ( \frac{x}{12} ) ^{2}}))

簡素化

\frac{1}{12} (arctan (\frac{x}{12}) - \frac{1}{2} (arctan (\frac{x}{12}) + \frac{\frac{x}{12}}{1 + ( \frac{x}{12} ) ^{2}})) : \frac{1}{12} (\frac{1}{2} arctan (\frac{x}{12}) - \frac{6 x}{144 + x ^{2}})

= \frac{1}{12} (\frac{1}{2} arctan (\frac{x}{12}) - \frac{6 x}{144 + x ^{2}})

定数を解答に追加する

= \frac{1}{12} (\frac{1}{2} arctan (\frac{x}{12}) - \frac{6 x}{144 + x ^{2}}) + C

グラフ

人気の例

limit as x approaches 0 of arctan(5/x)

x \to 0 lim (arctan (\frac{5}{x}))

integral of 6cos(x)

\int 6 cos (x) d x

-6y^{''}+54y^'=0,y(2)=1,y^'(-1)=1

- 6 y^{^{''}} + 54 y^{^{'}} = 0, y (2) = 1, y^{^{'}} (- 1) = 1

integral from 0 to 1 of 3x

\int_{0}^{1} 3 x d x

(dy)/(dx)= 2/(sqrt(16-x^2))

\frac{d y}{d x} = \frac{2}{16 - x ^{2}}