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

The integral from 1 to 2 of 1/(x^2+3x) is (3ln(2)-ln(5))/3

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

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

\frac{3 ln ( 2 ) - ln ( 5 )}{3}

Decimal Notation

0.15666 \dots

Solution steps

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

Take the partial fraction of

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

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

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

\int_{1}^{2} \frac{1}{3 ( x + 3 )} d x = \frac{1}{3} (ln (5) - 2 ln (2))

= \frac{1}{3} ln (2) - \frac{1}{3} (ln (5) - 2 ln (2))

Simplify

\frac{1}{3} ln (2) - \frac{1}{3} (ln (5) - 2 ln (2)) : \frac{3 ln ( 2 ) - ln ( 5 )}{3}

= \frac{3 ln ( 2 ) - ln ( 5 )}{3}

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What is the integral from 1 to 2 of 1/(x^2+3x) ?
The integral from 1 to 2 of 1/(x^2+3x) is (3ln(2)-ln(5))/3