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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

y^{^{'''}} - 2 y^{^{''}} - 4 y^{^{'}} + 8 y = 6 x e^{2 x}

f (x) = \frac{1}{x + \frac{1}{x}}

\int \frac{13}{( x - 7 ) ( x + 6 )} d x

\int e^{- \frac{x}{y}} d x

\frac{d}{d x} (a x a + x)

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