What is the integral from 0 to 1 of 1/((x+2)(x+1)) ?

The integral from 0 to 1 of 1/((x+2)(x+1)) is -ln(3)+2ln(2)

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integral from 0 to 1 of 1/((x+2)(x+1))

\int_{0}^{1} \frac{1}{( x + 2 ) ( x + 1 )} d x

- ln (3) + 2 ln (2)

Decimal

0.28768 \dots

Solution steps

\int_{0}^{1} \frac{1}{( x + 2 ) ( x + 1 )} d x

Take the partial fraction of

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

= \int_{0}^{1} - \frac{1}{x + 2} + \frac{1}{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

= - \int_{0}^{1} \frac{1}{x + 2} d x + \int_{0}^{1} \frac{1}{x + 1} d x

\int_{0}^{1} \frac{1}{x + 2} d x = ln (3) - ln (2)

\int_{0}^{1} \frac{1}{x + 1} d x = ln (2)

= - (ln (3) - ln (2)) + ln (2)

Simplify

= - ln (3) + 2 ln (2)

What is the integral from 0 to 1 of 1/((x+2)(x+1)) ?
The integral from 0 to 1 of 1/((x+2)(x+1)) is -ln(3)+2ln(2)