zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的微积分 >

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

解答

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

解答

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

求解步骤

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

对

x^{2} - x + 1

配方:

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

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

使用换元积分法

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

提出常数:

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

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

乘开

\frac{2 u + 1}{4 u ^{2} + 3} : \frac{2 u}{4 u ^{2} + 3} + \frac{1}{4 u ^{2} + 3}

使用积分加法定则:

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

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

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

代回

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

化简

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

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

解答补常数

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

作图

流行的例子

integral of 14sec^2(x)tan^6(x)

\int 14 sec^{2} (x) tan^{6} (x) d x

integral of t^2cos^2(t^3)sin(t^3)

\int t^{2} cos^{2} (t^{3}) sin (t^{3}) d t

integral of 8x+sin(x)

\int 8 x + sin (x) d x

integral of (4x^2)/(sqrt(1-x^6))

\int \frac{4 x ^{2}}{1 - x ^{6}} d x

integral of (e^{5x}+4)(e^{-2x}+15)

\int (e^{5 x} + 4) (e^{- 2 x} + 15) d x