What is the integral from 4 to 7 of 5/((x-2)(2x+1)) ?

The integral from 4 to 7 of 5/((x-2)(2x+1)) is ln(5)-ln(2)-ln(5/3)

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integral from 4 to 7 of 5/((x-2)(2x+1))

\int_{4}^{7} \frac{5}{( x - 2 ) ( 2 x + 1 )} d x

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

Decimal Notation

0.40546 \dots

Solution steps

\int_{4}^{7} \frac{5}{( x - 2 ) ( 2 x + 1 )} d x

Take the constant out:

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

= 5 \cdot \int_{4}^{7} \frac{1}{( x - 2 ) ( 2 x + 1 )} d x

Take the partial fraction of

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

= 5 \cdot \int_{4}^{7} \frac{1}{5 ( x - 2 )} - \frac{2}{5 ( 2 x + 1 )} 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_{4}^{7} \frac{1}{5 ( x - 2 )} d x - \int_{4}^{7} \frac{2}{5 ( 2 x + 1 )} d x)

\int_{4}^{7} \frac{1}{5 ( x - 2 )} d x = \frac{1}{5} (ln (5) - ln (2))

\int_{4}^{7} \frac{2}{5 ( 2 x + 1 )} d x = \frac{1}{5} ln (\frac{5}{3})

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

Simplify

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

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

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What is the integral from 4 to 7 of 5/((x-2)(2x+1)) ?
The integral from 4 to 7 of 5/((x-2)(2x+1)) is ln(5)-ln(2)-ln(5/3)