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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{2}{x} d x = 2 ln ∣ x ∣

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

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

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

Add a constant to the solution

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

4 x^{2} \frac{d y}{d x} = \frac{d y}{d x} + 4 x e^{- y}

\int_{0}^{1} π (y - y^{4}) d y

\int \frac{4 x + 6}{( x ^{2} + 3 x + 7 ) ^{3}} d x

\frac{x ^{5}}{8}

\frac{10}{s}

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