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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (4x^2-x+1)/(x^3-x^2+x-1)

\int \frac{4 x ^{2} - x + 1}{x ^{3} - x ^{2} + x - 1} d x

2 ln ∣ x - 1 ∣ + ln x^{2} + 1 + arctan (x) + C

Solution steps

\int \frac{4 x ^{2} - x + 1}{x ^{3} - x ^{2} + x - 1} d x

Expand

\frac{4 x ^{2} - x + 1}{x ^{3} - x ^{2} + x - 1} : \frac{4 x ^{2}}{x ^{3} - x ^{2} + x - 1} - \frac{x}{x ^{3} - x ^{2} + x - 1} + \frac{1}{x ^{3} - x ^{2} + 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{4 x ^{2}}{x ^{3} - x ^{2} + x - 1} d x - \int \frac{x}{x ^{3} - x ^{2} + x - 1} d x + \int \frac{1}{x ^{3} - x ^{2} + x - 1} d x

\int \frac{4 x ^{2}}{x ^{3} - x ^{2} + x - 1} d x = 4 (\frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{2} (\frac{1}{2} ln x^{2} + 1 + arctan (x)))

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

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

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

Simplify

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

= 2 ln ∣ x - 1 ∣ + ln x^{2} + 1 + arctan (x)

Add a constant to the solution

= 2 ln ∣ x - 1 ∣ + ln x^{2} + 1 + arctan (x) + C

\int ln ∣ x - 1 ∣ d x

f (x) = \frac{x}{1 - x ^{2}}

f (x) = \frac{2500 x}{ln ( 8 x + 8 )}

x y^{^{'}} + (x + \frac{1}{3}) y = e^{- 2 x} y^{- 2}

f (x) = x^{2} - x + 4, at x = 1

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