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

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

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

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

x - 4 ln ∣ x + 1 ∣ + 2 ln ∣ x ∣ + \frac{1}{x} + C

Solution steps

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

Expand

\frac{x ^{4} - 1}{x ^{2} ( x + 1 ) ^{2}} : \frac{x ^{4}}{x ^{2} ( x + 1 ) ^{2}} - \frac{1}{x ^{2} ( x + 1 ) ^{2}}

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

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

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

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

Simplify

x - 2 ln ∣ x + 1 ∣ - \frac{1}{x + 1} - (- 2 ln ∣ x ∣ - \frac{1}{x} + 2 ln ∣ x + 1 ∣ - \frac{1}{x + 1}) : x - 4 ln ∣ x + 1 ∣ + 2 ln ∣ x ∣ + \frac{1}{x}

= x - 4 ln ∣ x + 1 ∣ + 2 ln ∣ x ∣ + \frac{1}{x}

Add a constant to the solution

= x - 4 ln ∣ x + 1 ∣ + 2 ln ∣ x ∣ + \frac{1}{x} + C

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