What is the integral from 2 to 3 of 6/(2x^2+2x-4) ?

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

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integral from 2 to 3 of 6/(2x^2+2x-4)

\int_{2}^{3} \frac{6}{2 x ^{2} + 2 x - 4} d x

3 ln (2) - ln (5)

Decimal

0.47000 \dots

Solution steps

\int_{2}^{3} \frac{6}{2 x ^{2} + 2 x - 4} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int_{2}^{3} \frac{1}{2 x ^{2} + 2 x - 4} d x

Take the partial fraction of

\frac{1}{2 x ^{2} + 2 x - 4} : \frac{1}{6 ( x - 1 )} - \frac{1}{6 ( x + 2 )}

= 6 \cdot \int_{2}^{3} \frac{1}{6 ( x - 1 )} - \frac{1}{6 ( 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

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

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

\int_{2}^{3} \frac{1}{6 ( x + 2 )} d x = \frac{1}{6} (ln (5) - 2 ln (2))

= 6 (\frac{1}{6} ln (2) - \frac{1}{6} (ln (5) - 2 ln (2)))

Simplify

6 (\frac{1}{6} ln (2) - \frac{1}{6} (ln (5) - 2 ln (2))) : 3 ln (2) - ln (5)

= 3 ln (2) - ln (5)

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