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

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

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

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

- ln (3) + 2 ln (2)

Decimal Notation

0.28768 \dots

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

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