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

解

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

解

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

解答ステップ

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

u置換積分法を適用する

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

簡素化

\frac{3 + u}{2 + u} : \frac{3 + u}{2 + u}

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

u置換積分法を適用する

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

定数を除く:

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

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

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

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

(1 - v^{2}) 1 - v^{2} 1 - v^{2} = (1 - v^{2})^{2}

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

以下の部分分数を得る:

\frac{v ^{2}}{( 1 - v ^{2} ) ^{2}} : - \frac{1}{4 ( v + 1 )} + \frac{1}{4 ( v + 1 ) ^{2}} + \frac{1}{4 ( v - 1 )} + \frac{1}{4 ( v - 1 ) ^{2}}

= \frac{1}{2} (- 2 \cdot \int - \frac{1}{4 ( v + 1 )} + \frac{1}{4 ( v + 1 ) ^{2}} + \frac{1}{4 ( v - 1 )} + \frac{1}{4 ( v - 1 ) ^{2}} 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} (- 2 (- \int \frac{1}{4 ( v + 1 )} d v + \int \frac{1}{4 ( v + 1 ) ^{2}} d v + \int \frac{1}{4 ( v - 1 )} d v + \int \frac{1}{4 ( v - 1 ) ^{2}} d v))

\int \frac{1}{4 ( v + 1 )} d v = \frac{1}{4} ln ∣ v + 1 ∣

\int \frac{1}{4 ( v + 1 ) ^{2}} d v = - \frac{1}{4 ( v + 1 )}

\int \frac{1}{4 ( v - 1 )} d v = \frac{1}{4} ln ∣ v - 1 ∣

\int \frac{1}{4 ( v - 1 ) ^{2}} d v = - \frac{1}{4 ( v - 1 )}

= \frac{1}{2} (- 2 (- \frac{1}{4} ln ∣ v + 1 ∣ - \frac{1}{4 ( v + 1 )} + \frac{1}{4} ln ∣ v - 1 ∣ - \frac{1}{4 ( v - 1 )}))

代用を戻す

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

簡素化

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

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

定数を解答に追加する

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

グラフ

人気の例

integral of e^{t+3}

\int e^{t + 3} d t

integral of sqrt(e^{-2x)}

\int e^{- 2 x} d x

integral of (3-x)/(x-4)

\int \frac{3 - x}{x - 4} d x

integral of 3tsin(t)

\int 3 t sin (t) d t

integral of 1/3 csc(x)cot(x)

\int \frac{1}{3} csc (x) cot (x) d x