What is the integral of (8x)/(x^3+x^2-x-1) ?

The integral of (8x)/(x^3+x^2-x-1) is 8(-1/4 ln|x+1|-1/(2(x+1))+1/4 ln|x-1|)+C

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integral of (8x)/(x^3+x^2-x-1)

\int \frac{8 x}{x ^{3} + x ^{2} - x - 1} d x

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

Solution steps

\int \frac{8 x}{x ^{3} + x ^{2} - x - 1} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 \cdot \int \frac{x}{x ^{3} + x ^{2} - x - 1} d x

Take the partial fraction of

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

= 8 \cdot \int - \frac{1}{4 ( x + 1 )} + \frac{1}{2 ( x + 1 ) ^{2}} + \frac{1}{4 ( 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

= 8 (- \int \frac{1}{4 ( x + 1 )} d x + \int \frac{1}{2 ( x + 1 ) ^{2}} d x + \int \frac{1}{4 ( x - 1 )} d x)

\int \frac{1}{4 ( x + 1 )} d x = \frac{1}{4} ln ∣ x + 1 ∣

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

\int \frac{1}{4 ( x - 1 )} d x = \frac{1}{4} ln ∣ x - 1 ∣

= 8 (- \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{2 ( x + 1 )} + \frac{1}{4} ln ∣ x - 1 ∣)

Add a constant to the solution

= 8 (- \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{2 ( x + 1 )} + \frac{1}{4} ln ∣ x - 1 ∣) + C

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

\frac{\partial}{\partial y} (y^{2} e^{x y - 3})

x \to - 5 lim (\frac{3}{x + 5})

\int \frac{1}{- 2 y} d y

What is the integral of (8x)/(x^3+x^2-x-1) ?
The integral of (8x)/(x^3+x^2-x-1) is 8(-1/4 ln|x+1|-1/(2(x+1))+1/4 ln|x-1|)+C