What is the integral of ((x^2+1))/((x-9)(x-8)^2) ?

The integral of ((x^2+1))/((x-9)(x-8)^2) is 82ln|x-9|-81ln|x-8|+(65)/(x-8)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of ((x^2+1))/((x-9)(x-8)^2)

\int \frac{( x ^{2} + 1 )}{( x - 9 ) ( x - 8 ) ^{2}} d x

82 ln ∣ x - 9 ∣ - 81 ln ∣ x - 8 ∣ + \frac{65}{x - 8} + C

Solution steps

\int \frac{x ^{2} + 1}{( x - 9 ) ( x - 8 ) ^{2}} d x

Take the partial fraction of

\frac{x ^{2} + 1}{( x - 9 ) ( x - 8 ) ^{2}} : \frac{82}{x - 9} - \frac{81}{x - 8} - \frac{65}{( x - 8 ) ^{2}}

= \int \frac{82}{x - 9} - \frac{81}{x - 8} - \frac{65}{( x - 8 ) ^{2}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{82}{x - 9} d x - \int \frac{81}{x - 8} d x - \int \frac{65}{( x - 8 ) ^{2}} d x

\int \frac{82}{x - 9} d x = 82 ln ∣ x - 9 ∣

\int \frac{81}{x - 8} d x = 81 ln ∣ x - 8 ∣

\int \frac{65}{( x - 8 ) ^{2}} d x = - \frac{65}{x - 8}

= 82 ln ∣ x - 9 ∣ - 81 ln ∣ x - 8 ∣ - (- \frac{65}{x - 8})

Simplify

= 82 ln ∣ x - 9 ∣ - 81 ln ∣ x - 8 ∣ + \frac{65}{x - 8}

Add a constant to the solution

= 82 ln ∣ x - 9 ∣ - 81 ln ∣ x - 8 ∣ + \frac{65}{x - 8} + C

\int x cos (8 x^{2}) d x

f (x) = x^{2} - 2 x + 3, at x = 2

x \to 0 lim (Q (x))

\int \frac{7}{3 p - 4} d p

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

What is the integral of ((x^2+1))/((x-9)(x-8)^2) ?
The integral of ((x^2+1))/((x-9)(x-8)^2) is 82ln|x-9|-81ln|x-8|+(65)/(x-8)+C