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

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

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integral from 0 to 1 of 2/((x+1)(x+3))

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

- ln (2) + ln (3)

Decimal

0.40546 \dots

Solution steps

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

Take the constant out:

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

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

Take the partial fraction of

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

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

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

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

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