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

y^{^{''}} + \frac{1}{2} y = 0

f (x) = (5 x + 9) x^{2}

\frac{3}{x}, at x = \frac{1}{16}

\frac{d y}{d t} = y cos (3 t + 2)

x, 0, 0.5

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