What is the integral of ((x^2+1))/((x-10)(x-9)^2) ?

The integral of ((x^2+1))/((x-10)(x-9)^2) is 101ln|x-10|-100ln|x-9|+(82)/(x-9)+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-10)(x-9)^2)

\int \frac{( x ^{2} + 1 )}{( x - 10 ) ( x - 9 ) ^{2}} d x

101 ln ∣ x - 10 ∣ - 100 ln ∣ x - 9 ∣ + \frac{82}{x - 9} + C

Solution steps

\int \frac{x ^{2} + 1}{( x - 10 ) ( x - 9 ) ^{2}} d x

Take the partial fraction of

\frac{x ^{2} + 1}{( x - 10 ) ( x - 9 ) ^{2}} : \frac{101}{x - 10} - \frac{100}{x - 9} - \frac{82}{( x - 9 ) ^{2}}

= \int \frac{101}{x - 10} - \frac{100}{x - 9} - \frac{82}{( x - 9 ) ^{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{101}{x - 10} d x - \int \frac{100}{x - 9} d x - \int \frac{82}{( x - 9 ) ^{2}} d x

\int \frac{101}{x - 10} d x = 101 ln ∣ x - 10 ∣

\int \frac{100}{x - 9} d x = 100 ln ∣ x - 9 ∣

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

= 101 ln ∣ x - 10 ∣ - 100 ln ∣ x - 9 ∣ - (- \frac{82}{x - 9})

Simplify

= 101 ln ∣ x - 10 ∣ - 100 ln ∣ x - 9 ∣ + \frac{82}{x - 9}

Add a constant to the solution

= 101 ln ∣ x - 10 ∣ - 100 ln ∣ x - 9 ∣ + \frac{82}{x - 9} + C

f (x) = \frac{x ^{3}}{x ^{2} + 2}

(- 2, - 8), (3, 2)

(5.5) (4.2)

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

\frac{d}{d x} (\frac{x}{x ^{2} + 7})

What is the integral of ((x^2+1))/((x-10)(x-9)^2) ?
The integral of ((x^2+1))/((x-10)(x-9)^2) is 101ln|x-10|-100ln|x-9|+(82)/(x-9)+C