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 (10x-17)/(sqrt(x^2-6x-55))

解答

\int \frac{10 x - 17}{x ^{2} - 6 x - 55} d x

解答

10 x^{2} - 6 x - 55 + 13 ln (\frac{1}{8} x^{2} - 6 x - 55 + x - 3) + C

求解步骤

\int \frac{10 x - 17}{x ^{2} - 6 x - 55} d x

对

x^{2} - 6 x - 55

配方:

(x - 3)^{2} - 64

= \int \frac{10 x - 17}{( x - 3 ) ^{2} - 64} d x

使用换元积分法

= \int \frac{10 u + 13}{u ^{2} - 64} d u

乘开

\frac{10 u + 13}{u ^{2} - 64} : \frac{10 u}{u ^{2} - 64} + \frac{13}{u ^{2} - 64}

= \int \frac{10 u}{u ^{2} - 64} + \frac{13}{u ^{2} - 64} d u

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{10 u}{u ^{2} - 64} d u + \int \frac{13}{u ^{2} - 64} d u

\int \frac{10 u}{u ^{2} - 64} d u = 10 u^{2} - 64

\int \frac{13}{u ^{2} - 64} d u = 13 ln (\frac{1}{8} u^{2} - 64 + u)

= 10 u^{2} - 64 + 13 ln (\frac{1}{8} u^{2} - 64 + u)

u = x - 3

代回

= 10 (x - 3)^{2} - 64 + 13 ln (\frac{1}{8} (x - 3)^{2} - 64 + x - 3)

化简

10 (x - 3)^{2} - 64 + 13 ln (\frac{1}{8} (x - 3)^{2} - 64 + x - 3) : 10 x^{2} - 6 x - 55 + 13 ln (\frac{1}{8} x^{2} - 6 x - 55 + x - 3)

= 10 x^{2} - 6 x - 55 + 13 ln (\frac{1}{8} x^{2} - 6 x - 55 + x - 3)

解答补常数

= 10 x^{2} - 6 x - 55 + 13 ln (\frac{1}{8} x^{2} - 6 x - 55 + x - 3) + C

作图

流行的例子

integral of tan^4(4x)

\int tan^{4} (4 x) d x

derivative \sqrt[6]{x}^{ln(x)}

d er i v a t i v e 6 x^{l n (x)}

integral from 1 to 9 of pi(x-1)

\int_{1}^{9} π (x - 1) d x

tangent f(x)= 5/x ,(5,1)

t an g e n t f (x) = \frac{5}{x}, (5, 1)

derivative ((x-1))/(x^2+3)

d er i v a t i v e \frac{( x - 1 )}{x ^{2} + 3}