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

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

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

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

- ln (2) + ln (3)

Decimal

0.40546 \dots

Solution steps

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

Take the constant out:

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

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

Take the partial fraction of

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

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

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

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

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

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

Simplify

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

= - ln (2) + ln (3)

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