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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

= \int \frac{3}{x} - \frac{1}{x ^{2}} - \frac{3}{x + 1} - \frac{2}{( x + 1 ) ^{2}} 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{3}{x} d x - \int \frac{1}{x ^{2}} d x - \int \frac{3}{x + 1} d x - \int \frac{2}{( x + 1 ) ^{2}} d x

\int \frac{3}{x} d x = 3 ln ∣ x ∣

\int \frac{1}{x ^{2}} d x = - \frac{1}{x}

\int \frac{3}{x + 1} d x = 3 ln ∣ x + 1 ∣

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

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

Simplify

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

Add a constant to the solution

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

y^{2} d y + x^{3} d x = 0

f (x) = \frac{7}{x ^{2}} - 6 x^{2}, at x = 5

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

\int \frac{( 1 - x )}{e ^{x}} k d x

x \to 0 lim (+ (ln (x)))

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