What is the integral of (x^2+x-4)/(x-1) ?

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

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

3 x + \frac{1}{2} (x - 1)^{2} - 2 ln ∣ x - 1 ∣ - 3 + C

Solution steps

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

Expand

\frac{x ^{2} + x - 4}{x - 1} : \frac{x ^{2}}{x - 1} + \frac{x}{x - 1} - \frac{4}{x - 1}

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

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

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

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

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

Simplify

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

= 3 x + \frac{1}{2} (x - 1)^{2} - 2 ln ∣ x - 1 ∣ - 3

Add a constant to the solution

= 3 x + \frac{1}{2} (x - 1)^{2} - 2 ln ∣ x - 1 ∣ - 3 + C

\int \frac{cos ( z )}{1 + sin ( z )} d z

\int 0.2 x + 1 d x

sec^{2} (\frac{x}{2})

\int x (2 - x^{2})^{3} d x

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

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