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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

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

Add a constant to the solution

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

\int \frac{29 x ^{4} + 29 x ^{3} + 2 x + 1}{x ^{2} + x} d x

\frac{d}{d x} y = (sin (\frac{x}{2}) - cos (\frac{x}{2}))^{2}

\int \frac{5 - x}{2 x ^{2} + x - 1} d x

\frac{\partial}{\partial x} (a r cot (\frac{y}{x}))

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