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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

y^{^{'}} = 3 x + y, y (0) = 7

x \to 0 + lim (x^{2} csc^{2} (x))

x \frac{d y}{d x} = y - \frac{1}{y}

\frac{d}{d x} (\frac{x ^{4} - 5 x ^{3} + x}{x ^{2}})

\frac{d y}{d x} = \frac{e ^{x} + 1}{2 y}

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