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

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

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

Expand

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

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

= 3 (\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

3 (\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 e^{x} sin (y) d x

\frac{d}{d x} (y^{2} - 2 y \cdot cos (x))

(1 + cos (t)) y^{^{'}} = (1 + e - y) sin (t), y (0) = 0

(x - 2)^{2} (x - 1)

\frac{d v}{d t} = 0.64 - \frac{v}{29400} \cdot 8, v (0) = 0

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