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)/(sqrt(x^2+2x+2))

解

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

解

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

解答ステップ

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

平方完成する

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

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

u置換積分法を適用する

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

拡張

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

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

\int \frac{u ^{2}}{u ^{2} + 1} d u = \frac{1}{2} u u^{2} + 1 - \frac{1}{2} ln u + u^{2} + 1

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

\int \frac{1}{u ^{2} + 1} d u = ln u^{2} + 1 + u

= \frac{1}{2} u u^{2} + 1 - \frac{1}{2} ln u + u^{2} + 1 - 2 u^{2} + 1 + ln u^{2} + 1 + u

代用を戻す

u = x + 1

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

簡素化

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

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

定数を解答に追加する

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

グラフ

人気の例

(\partial)/(\partial x)(x+2y-1)

\frac{\partial}{\partial x} (x + 2 y - 1)

面積-x^2-3x,[-6,-1]

a re a - x^{2} - 3 x, [- 6, - 1]

(\partial)/(\partial y)(2y)

\frac{\partial}{\partial y} (2 y)

x y^{^{'}} + 4 y = 5 x

d/(dy)((x^2)/(x^2+y^2))

\frac{d}{d y} (\frac{x ^{2}}{x ^{2} + y ^{2}})