What is the integral of (7x^2+2x-7)/(x^3-x) ?

The integral of (7x^2+2x-7)/(x^3-x) is 7ln|x|-ln|x+1|+ln|x-1|+C

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integral of (7x^2+2x-7)/(x^3-x)

\int \frac{7 x ^{2} + 2 x - 7}{x ^{3} - x} d x

7 ln ∣ x ∣ - ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + C

Solution steps

\int \frac{7 x ^{2} + 2 x - 7}{x ^{3} - x} d x

Take the partial fraction of

\frac{7 x ^{2} + 2 x - 7}{x ^{3} - x} : \frac{7}{x} - \frac{1}{x + 1} + \frac{1}{x - 1}

= \int \frac{7}{x} - \frac{1}{x + 1} + \frac{1}{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

= \int \frac{7}{x} d x - \int \frac{1}{x + 1} d x + \int \frac{1}{x - 1} d x

\int \frac{7}{x} d x = 7 ln ∣ x ∣

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

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

= 7 ln ∣ x ∣ - ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣

Add a constant to the solution

= 7 ln ∣ x ∣ - ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + C

f (x) = 2 x^{2} + x - 1, at x = - 1

\frac{d y}{d x} + y = y^{2}

\int x \cdot e^{1 - x^{2}} d x

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

x^{l n (x)}

What is the integral of (7x^2+2x-7)/(x^3-x) ?
The integral of (7x^2+2x-7)/(x^3-x) is 7ln|x|-ln|x+1|+ln|x-1|+C