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+1)/(x^2-6x+8)

解答

\int \frac{x + 1}{x ^{2} - 6 x + 8} d x

解答

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

求解步骤

\int \frac{x + 1}{x ^{2} - 6 x + 8} d x

对

x^{2} - 6 x + 8

配方:

(x - 3)^{2} - 1

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

使用换元积分法

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

乘开

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

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

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

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

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

u = x - 3

代回

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

化简

\frac{1}{2} ln (x - 3)^{2} - 1 - 4 (\frac{1}{2} ln ∣ x - 3 + 1 ∣ - \frac{1}{2} ln ∣ x - 3 - 1 ∣) : \frac{1}{2} ln x^{2} - 6 x + 8 - 4 (\frac{1}{2} ln ∣ x - 2 ∣ - \frac{1}{2} ln ∣ x - 4 ∣)

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

解答补常数

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

作图

流行的例子

integral of (x^4)/(x^5-7)

\int \frac{x ^{4}}{x ^{5} - 7} d x

integral of xsin(x^2)cos(x^2)

\int x sin (x^{2}) cos (x^{2}) d x

integral of e^t+1

\int e^{t} + 1 d t

integral of (1+cos^2(θ))/(cos^2(θ))

\int \frac{1 + cos ^{2} ( θ )}{cos ^{2} ( θ )} d θ

integral of 1/(x(sqrt(4+x^2))^3)

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