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+2)/(x^2-7x+12)

解答

\int \frac{x + 2}{x ^{2} - 7 x + 12} d x

解答

2 (\frac{1}{4} ln 4 x^{2} - 28 x + 48 - \frac{11}{4} (ln ∣ 2 x - 6 ∣ - ln ∣ 2 x - 8 ∣)) + C

求解步骤

\int \frac{x + 2}{x ^{2} - 7 x + 12} d x

对

x^{2} - 7 x + 12

配方:

(x - \frac{7}{2})^{2} - \frac{1}{4}

= \int \frac{x + 2}{( x - \frac{7}{2} ) ^{2} - \frac{1}{4}} d x

使用换元积分法

= \int \frac{2 ( 2 u + 11 )}{4 u ^{2} - 1} 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 + 11}{4 u ^{2} - 1} d u

乘开

\frac{2 u + 11}{4 u ^{2} - 1} : \frac{2 u}{4 u ^{2} - 1} + \frac{11}{4 u ^{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

= 2 (\int \frac{2 u}{4 u ^{2} - 1} d u + \int \frac{11}{4 u ^{2} - 1} d u)

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

\int \frac{11}{4 u ^{2} - 1} d u = - \frac{11}{4} (ln ∣ 2 u + 1 ∣ - ln ∣ 2 u - 1 ∣)

= 2 (\frac{1}{4} ln 4 u^{2} - 1 - \frac{11}{4} (ln ∣ 2 u + 1 ∣ - ln ∣ 2 u - 1 ∣))

u = x - \frac{7}{2}

代回

= 2 (\frac{1}{4} ln 4 (x - \frac{7}{2})^{2} - 1 - \frac{11}{4} (ln 2 (x - \frac{7}{2}) + 1 - ln 2 (x - \frac{7}{2}) - 1))

化简

2 (\frac{1}{4} ln 4 (x - \frac{7}{2})^{2} - 1 - \frac{11}{4} (ln 2 (x - \frac{7}{2}) + 1 - ln 2 (x - \frac{7}{2}) - 1)) : 2 (\frac{1}{4} ln 4 x^{2} - 28 x + 48 - \frac{11}{4} (ln ∣ 2 x - 6 ∣ - ln ∣ 2 x - 8 ∣))

= 2 (\frac{1}{4} ln 4 x^{2} - 28 x + 48 - \frac{11}{4} (ln ∣ 2 x - 6 ∣ - ln ∣ 2 x - 8 ∣))

解答补常数

= 2 (\frac{1}{4} ln 4 x^{2} - 28 x + 48 - \frac{11}{4} (ln ∣ 2 x - 6 ∣ - ln ∣ 2 x - 8 ∣)) + C

作图

流行的例子

integral of sqrt(x)(x^2+1)

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

integral of x^3(7+x^4)^6

\int x^{3} (7 + x^{4})^{6} d x

integral of sec^2(x)sin^3(x)

\int sec^{2} (x) sin^{3} (x) d x

integral of (6^x+1)/(2^x)

\int \frac{6 ^{x} + 1}{2 ^{x}} d x

integral of x/(4x^2+3)

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