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

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

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

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

2 ln (2) - ln (3)

Decimal Notation

0.28768 \dots

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Simplify

= 2 ln (2) - ln (3)

f (x) = 17 tan (x), at x = \frac{π}{4}

5 x^{4} 25 - x^{2}

f (x) = (x + 2) 2 x - 3

\frac{\partial}{\partial y} (\frac{( x + y )}{( y + z )})

\frac{\partial}{\partial x} (\frac{x}{x ^{6} - y ^{3}})

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