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

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

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integral of (4x)/(x^3+x^2-x-1)

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

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

Solution steps

\int \frac{4 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

= 4 \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 )}

= 4 \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

= 4 (- \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 ∣

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

Simplify

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

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

Add a constant to the solution

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

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

x \to - 1 lim (- x^{2} + 2 x - 9)

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

in v erse l a pl a ce \frac{s}{s ^{2} + 4 s + 3}

\frac{\partial}{\partial x} (4 x^{0.75} + y)

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