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

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

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

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

ln ∣ x ∣ - ln ∣ x + 1 ∣ + C

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

= ln ∣ x ∣ - ln ∣ x + 1 ∣

Add a constant to the solution

= ln ∣ x ∣ - ln ∣ x + 1 ∣ + C

\int sin (4 x + 3) d x

\frac{d y}{d x} + y = x + 1

cos (2 x)

y^{^{'}} = \frac{y}{x} - 1

\int sin^{2} (x) \cdot cos^{2} (x) d x

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