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

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

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

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

\frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{x - 1} - \frac{1}{2 ( x - 1 )} + C

Solution steps

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

Expand

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

f (y) = arcsin (y^{2})

\int e^{8 x} (8) d x

\int_{1}^{\infty} \frac{x}{5 + x ^{2}} d x

5 (2 x (3 x^{2} - 1)^{2} + 12 x (3 x^{2} - 1))

- \frac{1}{2} e^{- 2 x}

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