What is the integral from 0 to 3 of 1/((x+2)(x+4)) ?

The integral from 0 to 3 of 1/((x+2)(x+4)) is (ln(5)+ln(2)-ln(7))/2

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integral from 0 to 3 of 1/((x+2)(x+4))

\int_{0}^{3} \frac{1}{( x + 2 ) ( x + 4 )} d x

\frac{ln ( 5 ) + ln ( 2 ) - ln ( 7 )}{2}

Decimal Notation

0.17833 \dots

Solution steps

\int_{0}^{3} \frac{1}{( x + 2 ) ( x + 4 )} d x

Take the partial fraction of

\frac{1}{( x + 2 ) ( x + 4 )} : \frac{1}{2 ( x + 2 )} - \frac{1}{2 ( x + 4 )}

= \int_{0}^{3} \frac{1}{2 ( x + 2 )} - \frac{1}{2 ( x + 4 )} 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}^{3} \frac{1}{2 ( x + 2 )} d x - \int_{0}^{3} \frac{1}{2 ( x + 4 )} d x

\int_{0}^{3} \frac{1}{2 ( x + 2 )} d x = \frac{1}{2} (ln (5) - ln (2))

\int_{0}^{3} \frac{1}{2 ( x + 4 )} d x = \frac{1}{2} ln (7) - ln (2)

= \frac{1}{2} (ln (5) - ln (2)) - (\frac{1}{2} ln (7) - ln (2))

Simplify

\frac{1}{2} (ln (5) - ln (2)) - (\frac{1}{2} ln (7) - ln (2)) : \frac{ln ( 5 ) + ln ( 2 ) - ln ( 7 )}{2}

= \frac{ln ( 5 ) + ln ( 2 ) - ln ( 7 )}{2}

\frac{10 s}{s ^{2} - 25}

x \to - 5 lim (- 1 x)

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

\int x e^{2} x d x

\int \frac{x ^{3} + 5}{x ^{2}} d x

What is the integral from 0 to 3 of 1/((x+2)(x+4)) ?
The integral from 0 to 3 of 1/((x+2)(x+4)) is (ln(5)+ln(2)-ln(7))/2