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/((x^2+4x+5)^{3/2)}

解

\int \frac{x}{( x ^{2} + 4 x + 5 ) ^{\frac{3}{2}}} d x

解

- \frac{1}{( x ^{2} + 4 x + 5 ) ^{\frac{1}{2}}} - \frac{2 ( x + 2 )}{( x ^{2} + 4 x + 5 ) ^{\frac{1}{2}}} + C

解答ステップ

\int \frac{x}{( x ^{2} + 4 x + 5 ) ^{\frac{3}{2}}} d x

平方完成する

x^{2} + 4 x + 5 : (x + 2)^{2} + 1

= \int \frac{x}{( ( x + 2 ) ^{2} + 1 ) ^{\frac{3}{2}}} d x

u置換積分法を適用する

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

拡張

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

= \int \frac{u}{( u ^{2} + 1 ) ^{\frac{3}{2}}} - \frac{2}{( u ^{2} + 1 ) ^{\frac{3}{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

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

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

\int \frac{2}{( u ^{2} + 1 ) ^{\frac{3}{2}}} d u = \frac{2 u}{( 1 + u ^{2} ) ^{\frac{1}{2}}}

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

代用を戻す

u = x + 2

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

簡素化

- \frac{1}{( ( x + 2 ) ^{2} + 1 ) ^{\frac{1}{2}}} - \frac{2 ( x + 2 )}{( 1 + ( x + 2 ) ^{2} ) ^{\frac{1}{2}}} : - \frac{1}{( x ^{2} + 4 x + 5 ) ^{\frac{1}{2}}} - \frac{2 ( x + 2 )}{( x ^{2} + 4 x + 5 ) ^{\frac{1}{2}}}

= - \frac{1}{( x ^{2} + 4 x + 5 ) ^{\frac{1}{2}}} - \frac{2 ( x + 2 )}{( x ^{2} + 4 x + 5 ) ^{\frac{1}{2}}}

定数を解答に追加する

= - \frac{1}{( x ^{2} + 4 x + 5 ) ^{\frac{1}{2}}} - \frac{2 ( x + 2 )}{( x ^{2} + 4 x + 5 ) ^{\frac{1}{2}}} + C

グラフ

人気の例

(dy)/(dx)=((x-4))/((y+5))

\frac{d y}{d x} = \frac{( x - 4 )}{( y + 5 )}

derivative ax^2+bx+c

d er i v a t i v e a x^{2} + b x + c

derivative of (6x/(1-tan(x)))

\frac{d}{d x} (\frac{6 x}{1 - tan ( x )})

integral from 0 to pi/4 of 6sec^2(x)

\int_{0}^{\frac{π}{4}} 6 sec^{2} (x) d x

integral of (sqrt(y^5-1))/y

\int \frac{y ^{5} - 1}{y} d y