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

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

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integral from 3 to u of 1/(x+x^2)

\int_{3}^{u} \frac{1}{x + x ^{2}} d x

ln ∣ u ∣ - ln (3) - ln ∣ u + 1 ∣ + 2 ln (2)

Solution steps

\int_{3}^{u} \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_{3}^{u} \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_{3}^{u} \frac{1}{x} d x - \int_{3}^{u} \frac{1}{x + 1} d x

\int_{3}^{u} \frac{1}{x} d x = ln ∣ u ∣ - ln (3)

\int_{3}^{u} \frac{1}{x + 1} d x = ln ∣ u + 1 ∣ - 2 ln (2)

= ln ∣ u ∣ - ln (3) - (ln ∣ u + 1 ∣ - 2 ln (2))

Simplify

= ln ∣ u ∣ - ln (3) - ln ∣ u + 1 ∣ + 2 ln (2)

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