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

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

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

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

ln (7) - ln (3) - ln (2)

Decimal

0.15415 \dots

Solution steps

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

Take the partial fraction of

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

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

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

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

= ln (7) - ln (3) - ln (2)

\int_{4}^{6} x^{2} + 2 x - 5 d x

\int_{0}^{1} cos (x) - 0.5 d x

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\int_{0}^{1} π e^{2 x} d x

\int_{x}^{π} cos (t^{2}) d t

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