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

The integral of (x^2+4x)/(x^3-x^2+x-1) is 5/2 ln|x-1|-3/4 ln|x^2+1|+5/2 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 (x^2+4x)/(x^3-x^2+x-1)

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

\frac{5}{2} ln ∣ x - 1 ∣ - \frac{3}{4} ln x^{2} + 1 + \frac{5}{2} arctan (x) + C

Solution steps

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

Expand

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

\int \frac{x ^{2}}{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{4 x}{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)))

= \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)))

Simplify

\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{5}{2} ln ∣ x - 1 ∣ - \frac{3}{4} ln x^{2} + 1 + \frac{5}{2} arctan (x)

= \frac{5}{2} ln ∣ x - 1 ∣ - \frac{3}{4} ln x^{2} + 1 + \frac{5}{2} arctan (x)

Add a constant to the solution

= \frac{5}{2} ln ∣ x - 1 ∣ - \frac{3}{4} ln x^{2} + 1 + \frac{5}{2} arctan (x) + C

\int \frac{a ^{3}}{x ^{2} + a ^{2}} d x

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

\int sin (18 x) cos (15 x) d x

\int \frac{1}{3 x - 9} d x

\int (\frac{8 x ^{3} - 4 x ^{2} - 3 x}{6 x}) d x

What is the integral of (x^2+4x)/(x^3-x^2+x-1) ?
The integral of (x^2+4x)/(x^3-x^2+x-1) is 5/2 ln|x-1|-3/4 ln|x^2+1|+5/2 arctan(x)+C