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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

\int sec (3 x) tan (3 x) d x

\int (\frac{3}{x} - \frac{10}{x ^{6}}) d x

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

\int 1 + t d t

\frac{d}{d x} (\frac{( x + 30 ) x ^{2} + 100}{x})

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