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-1))/(x^2+x+1)

解

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

解

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

解答ステップ

\int \frac{x - 1}{x ^{2} + x + 1} d x

平方完成する

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

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

u置換積分法を適用する

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

定数を除く:

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

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

拡張

\frac{2 u - 3}{4 u ^{2} + 3} : \frac{2 u}{4 u ^{2} + 3} - \frac{3}{4 u ^{2} + 3}

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

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

\int \frac{2 u}{4 u ^{2} + 3} d u = \frac{1}{4} ln 4 u^{2} + 3

\int \frac{3}{4 u ^{2} + 3} d u = \frac{3}{2} arctan (\frac{2}{3} u)

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

代用を戻す

u = x + \frac{1}{2}

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

簡素化

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

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

定数を解答に追加する

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

グラフ

人気の例

integral from 0 to 1 of x(sqrt(x)-x)

\int_{0}^{1} x (x - x) d x

derivative of ln(sqrt(x-2))

\frac{d}{d x} (ln (x - 2))

integral from 0 to pi of (5e^x+5sin(x))

\int_{0}^{π} (5 e^{x} + 5 sin (x)) d x

derivative f(x)= 1/(2x)

d er i v a t i v e f (x) = \frac{1}{2 x}

(\partial)/(\partial x)(sin(xy^2))

\frac{\partial}{\partial x} (sin (x y^{2}))