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

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

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

integral

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

ln ∣ x ∣ - ln ∣ x + 1 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{1}{x ^{2} + x} : \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{x}{( 5 x + 2 ) ^{3}}

\frac{1}{( x ^{2} + 4 )}

\frac{1}{( a ^{2} - x ^{2} ) ^{\frac{1}{2}}}

\frac{e ^{1}}{x}

1 + tan^{2} (x)

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