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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

= 3 ln ∣ x + 1 ∣ + \frac{2}{x + 1} + ln ∣ x - 1 ∣ - ln ∣ x + 1 ∣ - \frac{2}{x + 1} + ln ∣ x - 1 ∣ - (- ln ∣ x + 1 ∣ + \frac{2}{x + 1} + ln ∣ x - 1 ∣)

Simplify

3 ln ∣ x + 1 ∣ + \frac{2}{x + 1} + ln ∣ x - 1 ∣ - ln ∣ x + 1 ∣ - \frac{2}{x + 1} + ln ∣ x - 1 ∣ - (- ln ∣ x + 1 ∣ + \frac{2}{x + 1} + ln ∣ x - 1 ∣) : 3 ln ∣ x + 1 ∣ - \frac{2}{x + 1} + ln ∣ x - 1 ∣

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

Add a constant to the solution

= 3 ln ∣ x + 1 ∣ - \frac{2}{x + 1} + ln ∣ x - 1 ∣ + C

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

x, at x = 1

\int x (x - 3) d x

\frac{d}{d x} (3 - 3 x^{2})

f (x) = - \frac{7}{( x + 4 ) ^{2}}

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