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

The integral of (x^2+2x+1)/((x-2)^2(x-1)) is -3ln|x-2|-9/(x-2)+4ln|x-1|+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+2x+1)/((x-2)^2(x-1))

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

- 3 ln ∣ x - 2 ∣ - \frac{9}{x - 2} + 4 ln ∣ x - 1 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{x ^{2} + 2 x + 1}{( x - 2 ) ^{2} ( x - 1 )} : - \frac{3}{x - 2} + \frac{9}{( x - 2 ) ^{2}} + \frac{4}{x - 1}

= \int - \frac{3}{x - 2} + \frac{9}{( x - 2 ) ^{2}} + \frac{4}{x - 1} 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{3}{x - 2} d x + \int \frac{9}{( x - 2 ) ^{2}} d x + \int \frac{4}{x - 1} d x

\int \frac{3}{x - 2} d x = 3 ln ∣ x - 2 ∣

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

\int \frac{4}{x - 1} d x = 4 ln ∣ x - 1 ∣

= - 3 ln ∣ x - 2 ∣ - \frac{9}{x - 2} + 4 ln ∣ x - 1 ∣

Add a constant to the solution

= - 3 ln ∣ x - 2 ∣ - \frac{9}{x - 2} + 4 ln ∣ x - 1 ∣ + C

\int x (x^{2} + 5)^{9} d x

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

\int \frac{cos ^{3} ( 5 x )}{sin ^{4} ( 5 x )} d x

\int 6 x y^{2} - y^{3} d x

What is the integral of (x^2+2x+1)/((x-2)^2(x-1)) ?
The integral of (x^2+2x+1)/((x-2)^2(x-1)) is -3ln|x-2|-9/(x-2)+4ln|x-1|+C