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

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

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

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

2 ln (2) - ln (3)

Decimal Notation

0.28768 \dots

Solution steps

\int_{1}^{2} \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_{1}^{2} \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_{1}^{2} \frac{1}{x} d x - \int_{1}^{2} \frac{1}{x + 1} d x

\int_{1}^{2} \frac{1}{x} d x = ln (2)

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

= ln (2) - (ln (3) - ln (2))

Simplify

= 2 ln (2) - ln (3)

\int_{- 5}^{- 3} \frac{1}{x ^{2} - 9} d x

\int_{1}^{3} ln (x) + sin (x) + 8 d x

\int_{0}^{2} x (x - 2)^{2} d x

\int_{1}^{2} \frac{1}{x ^{3} + 8} d x

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