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))

integral

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

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

\frac{2}{t}

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

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

\frac{ln ( x )}{( 1 + x ^{2} )}

\frac{( x + 5 ) ^{3}}{2}

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