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

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

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

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

ln (3) - ln (2)

Decimal Notation

0.40546 \dots

Solution steps

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

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

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

= ln (3) - ln (2)

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

\int_{1}^{e} \frac{1}{x + x ln ( x )} d x

\int_{1}^{2} x^{\frac{1}{8}} ln (x) d x

\int_{- 1}^{1} x - 1 d x

\int_{0}^{1 - y} 1 d z

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