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

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

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

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

3 ln (2) - ln (5)

Decimal Notation

0.47000 \dots

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Take the partial fraction of

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

= 3 \cdot \int_{2}^{3} \frac{1}{3 ( x - 1 )} - \frac{1}{3 ( 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

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

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

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

= 3 (\frac{1}{3} ln (2) - \frac{1}{3} (ln (5) - 2 ln (2)))

Simplify

3 (\frac{1}{3} ln (2) - \frac{1}{3} (ln (5) - 2 ln (2))) : 3 ln (2) - ln (5)

= 3 ln (2) - ln (5)

\frac{\partial}{\partial x} (3 x - 2 y)

\int_{1}^{6} \frac{1}{( 6 - x ) ^{\frac{2}{3}}} d x

\int \frac{3}{2} (t + 1)^{\frac{1}{2}} d t

\frac{1}{t} \frac{d y}{d t} - \frac{1}{1 + t ^{2}} = 0

\frac{x ^{3} + 7}{x}

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