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 (2x-3)/(x^2+10x+29)

解答

\int \frac{2 x - 3}{x ^{2} + 10 x + 29} d x

解答

ln \frac{1}{4} (x + 5)^{2} + 1 - \frac{13}{2} arctan (\frac{1}{2} (x + 5)) + C

求解步骤

\int \frac{2 x - 3}{x ^{2} + 10 x + 29} d x

对

x^{2} + 10 x + 29

配方:

(x + 5)^{2} + 4

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

使用换元积分法

= \int \frac{2 u - 13}{u ^{2} + 4} d u

使用换元积分法

= \int \frac{4 v - 13}{2 ( v ^{2} + 1 )} d v

提出常数:

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

= \frac{1}{2} \cdot \int \frac{4 v - 13}{v ^{2} + 1} d v

乘开

\frac{4 v - 13}{v ^{2} + 1} : \frac{4 v}{v ^{2} + 1} - \frac{13}{v ^{2} + 1}

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

= \frac{1}{2} (\int \frac{4 v}{v ^{2} + 1} d v - \int \frac{13}{v ^{2} + 1} d v)

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

\int \frac{13}{v ^{2} + 1} d v = 13 arctan (v)

= \frac{1}{2} (2 ln v^{2} + 1 - 13 arctan (v))

代回

= \frac{1}{2} (2 ln (\frac{x + 5}{2})^{2} + 1 - 13 arctan (\frac{x + 5}{2}))

化简

\frac{1}{2} (2 ln (\frac{x + 5}{2})^{2} + 1 - 13 arctan (\frac{x + 5}{2})) : ln \frac{1}{4} (x + 5)^{2} + 1 - \frac{13}{2} arctan (\frac{1}{2} (x + 5))

= ln \frac{1}{4} (x + 5)^{2} + 1 - \frac{13}{2} arctan (\frac{1}{2} (x + 5))

解答补常数

= ln \frac{1}{4} (x + 5)^{2} + 1 - \frac{13}{2} arctan (\frac{1}{2} (x + 5)) + C

作图

流行的例子

derivative of x+cot(x/2)

\frac{d}{d x} (x + cot (\frac{x}{2}))

(\partial)/(\partial y)(10xye^{-x^2-y^2})

\frac{\partial}{\partial y} (10 x y e^{- x^{2} - y^{2}})

derivative of-3cot(1-x^2)

\frac{d}{d x} (- 3 cot (1 - x^{2}))

limit as x approaches 0 of 3-1/x

x \to 0 lim (3 - \frac{1}{x})

f (x) = π^{x}