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

y x = c

\frac{d}{d x} (\frac{4}{2 x + 5})

ln (\frac{x - 5}{x - 4})

\frac{d}{d x} (- x e^{- \frac{x ^{2}}{2}})

f^{^{'}} (x) = 2 sec (x) + 3 tan (x) - tan (\frac{π}{3})

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