What is the integral of (x^2-8x+7)/((x^2-7x+6)^2) ?

The integral of (x^2-8x+7)/((x^2-7x+6)^2) is -6/25 ln|x-1|+6/25 ln|x-6|+1/(5(x-6))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^2-8x+7)/((x^2-7x+6)^2)

\int \frac{x ^{2} - 8 x + 7}{( x ^{2} - 7 x + 6 ) ^{2}} d x

- \frac{6}{25} ln ∣ x - 1 ∣ + \frac{6}{25} ln ∣ x - 6 ∣ + \frac{1}{5 ( x - 6 )} + C

Solution steps

\int \frac{x ^{2} - 8 x + 7}{( x ^{2} - 7 x + 6 ) ^{2}} d x

Take the partial fraction of

\frac{x ^{2} - 8 x + 7}{( x ^{2} - 7 x + 6 ) ^{2}} : - \frac{6}{25 ( x - 1 )} + \frac{6}{25 ( x - 6 )} - \frac{1}{5 ( x - 6 ) ^{2}}

= \int - \frac{6}{25 ( x - 1 )} + \frac{6}{25 ( x - 6 )} - \frac{1}{5 ( x - 6 ) ^{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{6}{25 ( x - 1 )} d x + \int \frac{6}{25 ( x - 6 )} d x - \int \frac{1}{5 ( x - 6 ) ^{2}} d x

\int \frac{6}{25 ( x - 1 )} d x = \frac{6}{25} ln ∣ x - 1 ∣

\int \frac{6}{25 ( x - 6 )} d x = \frac{6}{25} ln ∣ x - 6 ∣

\int \frac{1}{5 ( x - 6 ) ^{2}} d x = - \frac{1}{5 ( x - 6 )}

= - \frac{6}{25} ln ∣ x - 1 ∣ + \frac{6}{25} ln ∣ x - 6 ∣ - (- \frac{1}{5 ( x - 6 )})

Simplify

= - \frac{6}{25} ln ∣ x - 1 ∣ + \frac{6}{25} ln ∣ x - 6 ∣ + \frac{1}{5 ( x - 6 )}

Add a constant to the solution

= - \frac{6}{25} ln ∣ x - 1 ∣ + \frac{6}{25} ln ∣ x - 6 ∣ + \frac{1}{5 ( x - 6 )} + C

\int 9^{1 - 4 x} d x

\int (sec (x) + cos (x))^{2} d x

\int 2 y y d y

\int \frac{sin ^{7} ( x )}{cos ^{3} ( x )} d x

\int (x^{3} + 3 x) d x

What is the integral of (x^2-8x+7)/((x^2-7x+6)^2) ?
The integral of (x^2-8x+7)/((x^2-7x+6)^2) is -6/25 ln|x-1|+6/25 ln|x-6|+1/(5(x-6))+C