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

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

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

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

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

Decimal Notation

0.05753 \dots

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Simplify

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

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

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

\int_{0}^{2} (3 x^{2}) d x

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

\int_{0}^{1} x^{2} 1 + 4 x^{2} d x

\int_{0}^{5} 5 ln (x) d x

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