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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int 1 d x = x

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

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

Simplify

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

Add a constant to the solution

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

\frac{d}{d y} (ln (y^{2}))

\int (\frac{x ^{3} + 7}{x ^{2}}) d x

\int_{- 3}^{0} ∣ x + 2 ∣ d x

\frac{d}{d x} (\frac{1}{arcsin ( x )})

\frac{\partial}{\partial x} (\frac{x}{( x + y )})

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