What is the integral of (x^2-3x+1)/(x+1) ?

The integral of (x^2-3x+1)/(x+1) is -5x+5ln|x+1|+1/2 (x+1)^2-5+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^2-3x+1)/(x+1)

\int \frac{x ^{2} - 3 x + 1}{x + 1} d x

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

Solution steps

\int \frac{x ^{2} - 3 x + 1}{x + 1} d x

Expand

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

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

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

Simplify

\frac{( x + 1 ) ^{2}}{2} - 2 (x + 1) + ln ∣ x + 1 ∣ - 3 (x + 1 - ln ∣ x + 1 ∣) + ln ∣ x + 1 ∣ : - 5 x + 5 ln ∣ x + 1 ∣ + \frac{1}{2} (x + 1)^{2} - 5

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

Add a constant to the solution

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

\frac{\partial}{\partial y} (3 x e^{- 2 x y})

(\frac{d y}{d x}) = y - y

\int \frac{1}{x ^{2} + 4 x + 8} d x

\frac{d}{d x} ((3 x - 5)^{4})

5 e^{2 x} \frac{d y}{d x} = - 10 \frac{x}{y ^{2}}

What is the integral of (x^2-3x+1)/(x+1) ?
The integral of (x^2-3x+1)/(x+1) is -5x+5ln|x+1|+1/2 (x+1)^2-5+C