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

The integral from 3 to 5 of 5/((2x+1)(x-2)) is -ln(11)+ln(7)+ln(3)

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integral from 3 to 5 of 5/((2x+1)(x-2))

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

- ln (11) + ln (7) + ln (3)

Decimal

0.64662 \dots

Solution steps

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

Take the constant out:

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

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

Take the partial fraction of

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

= 5 \cdot \int_{3}^{5} - \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

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

\int_{3}^{5} \frac{2}{5 ( 2 x + 1 )} d x = \frac{1}{5} (ln (11) - ln (7))

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

= 5 (- \frac{1}{5} (ln (11) - ln (7)) + \frac{1}{5} ln (3))

Simplify

5 (- \frac{1}{5} (ln (11) - ln (7)) + \frac{1}{5} ln (3)) : - ln (11) + ln (7) + ln (3)

= - ln (11) + ln (7) + ln (3)

What is the integral from 3 to 5 of 5/((2x+1)(x-2)) ?
The integral from 3 to 5 of 5/((2x+1)(x-2)) is -ln(11)+ln(7)+ln(3)