What is the integral from 0 to 1 of x/(2x^2+5x+2) ?

The integral from 0 to 1 of x/(2x^2+5x+2) is (3ln(3)-4ln(2))/6

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integral from 0 to 1 of x/(2x^2+5x+2)

\int_{0}^{1} \frac{x}{2 x ^{2} + 5 x + 2} d x

\frac{3 ln ( 3 ) - 4 ln ( 2 )}{6}

Decimal

0.08720 \dots

Solution steps

\int_{0}^{1} \frac{x}{2 x ^{2} + 5 x + 2} d x

Take the partial fraction of

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

= \int_{0}^{1} - \frac{1}{3 ( 2 x + 1 )} + \frac{2}{3 ( x + 2 )} 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_{0}^{1} \frac{1}{3 ( 2 x + 1 )} d x + \int_{0}^{1} \frac{2}{3 ( x + 2 )} d x

\int_{0}^{1} \frac{1}{3 ( 2 x + 1 )} d x = \frac{1}{6} ln (3)

\int_{0}^{1} \frac{2}{3 ( x + 2 )} d x = \frac{2}{3} (ln (3) - ln (2))

= - \frac{1}{6} ln (3) + \frac{2}{3} (ln (3) - ln (2))

Simplify

- \frac{1}{6} ln (3) + \frac{2}{3} (ln (3) - ln (2)) : \frac{3 ln ( 3 ) - 4 ln ( 2 )}{6}

= \frac{3 ln ( 3 ) - 4 ln ( 2 )}{6}

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What is the integral from 0 to 1 of x/(2x^2+5x+2) ?
The integral from 0 to 1 of x/(2x^2+5x+2) is (3ln(3)-4ln(2))/6