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

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

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

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

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

Decimal

0.11778 \dots

Solution steps

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

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

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

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

Simplify

- ln (2) + 2 (ln (3) - ln (2)) : - 3 ln (2) + 2 ln (3)

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

\int_{- 1}^{2} 1 - x d x

\int_{3}^{4} x^{3} d x

\int_{2}^{3} \frac{1}{2 y - 3} d y

\int_{1}^{2} \frac{1}{2} x^{2} d x

\int_{- 1}^{1} x^{2} (x^{3} + 3)^{4} d x

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