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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (x^{2} - 5 x y + 3 y^{2})

\int 25 arctan (x) d x

\frac{d ^{2} y}{d x ^{2}} + 20 \frac{d y}{d x} + 200 y = 300

\int_{1}^{\infty} 3 e^{- x} d x

7 x^{2} - 3 x

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