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

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

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

x + 2 ln ∣ x - 1 ∣ - ln ∣ x ∣ - 1 + C

Solution steps

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

Expand

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

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

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

= x - 1 + ln ∣ x - 1 ∣ - ln ∣ x ∣ + ln ∣ x - 1 ∣

Simplify

= x + 2 ln ∣ x - 1 ∣ - ln ∣ x ∣ - 1

Add a constant to the solution

= x + 2 ln ∣ x - 1 ∣ - ln ∣ x ∣ - 1 + C

\frac{\partial}{\partial y} (8 x ln (x y))

t^{2} \cdot y^{^{'}} + t \cdot y = 5

\frac{d}{d x} (e^{x l n (x)})

x \to 0 lim ((e^{x} + x)^{\frac{m}{x}})

\int \frac{7}{9 - 49 x ^{2}} d x

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