What is the integral of (7x^2-9x+6)/((x-1)(x^2+1)) ?

The integral of (7x^2-9x+6)/((x-1)(x^2+1)) is 2ln|x-1|+5/2 ln|x^2+1|-4arctan(x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (7x^2-9x+6)/((x-1)(x^2+1))

\int \frac{7 x ^{2} - 9 x + 6}{( x - 1 ) ( x ^{2} + 1 )} d x

2 ln ∣ x - 1 ∣ + \frac{5}{2} ln x^{2} + 1 - 4 arctan (x) + C

Solution steps

\int \frac{7 x ^{2} - 9 x + 6}{( x - 1 ) ( x ^{2} + 1 )} d x

Take the partial fraction of

\frac{7 x ^{2} - 9 x + 6}{( x - 1 ) ( x ^{2} + 1 )} : \frac{2}{x - 1} + \frac{5 x - 4}{x ^{2} + 1}

= \int \frac{2}{x - 1} + \frac{5 x - 4}{x ^{2} + 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{2}{x - 1} d x + \int \frac{5 x - 4}{x ^{2} + 1} d x

\int \frac{2}{x - 1} d x = 2 ln ∣ x - 1 ∣

\int \frac{5 x - 4}{x ^{2} + 1} d x = \frac{5}{2} ln x^{2} + 1 - 4 arctan (x)

= 2 ln ∣ x - 1 ∣ + \frac{5}{2} ln x^{2} + 1 - 4 arctan (x)

Add a constant to the solution

= 2 ln ∣ x - 1 ∣ + \frac{5}{2} ln x^{2} + 1 - 4 arctan (x) + C

\int \frac{ln ( 4 x ^{2} )}{x} d x

\int 19 x sin (x) cos (x) d x

\int x e x^{2} - 1 d x

\int sin^{4} (7 θ) d θ

\int \frac{x ^{4}}{x ^{2} + 225} d x

What is the integral of (7x^2-9x+6)/((x-1)(x^2+1)) ?
The integral of (7x^2-9x+6)/((x-1)(x^2+1)) is 2ln|x-1|+5/2 ln|x^2+1|-4arctan(x)+C