Can someone explain how to do this please?

Mathematics

1 answer:

Rufina [12.5K]3 years ago

3 0

for the columns where is less than or equal zero, use the first line of the function to calculate the value of f(x)

${ ( \frac{1}{3} ) }^{ - 2} - 1 = 9 - 1 = 8 \\ { ( \frac{1}{3} ) }^{ - 1} - 1 = 3 - 1 = 2 \\ { ( \frac{1}{3} ) }^{ 0} - 1 = 1 - 1 = 0$

for every x greater than zero, use the second line to determine the value

${3}^{1} - 2 = 3 - 2 = 1 \\ {3}^{2} - 2 = 9 - 2 = 7$

hope this helps and gives some insight what to do.

if there are open questions left, feel free to ask them in the comments

have a beautiful day

-Alex

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Offering 15 points for a good and explained answer.

Romashka [77]

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Ierofanga [76]

$\dfrac{x^2+x+1}{(x+1)^2(x+2)}=\dfrac{a_1x+a_0}{(x+1)^2}+\dfrac b{x+2}$
$\implies\dfrac{x^2+x+1}{(x+1)^2(x+2)}=\dfrac{(a_1x+a_0)(x+2)+b(x+1)^2}{(x+1)^2(x+2)}$
$\implies x^2+x+1=(a_1+b)x^2+(2a_1+a_0+2b)x+(2a_0+b)$
$\implies\begin{cases}a_1+b=1\\2a_1+a_0+2b=1\\2a_0+b=1\end{cases}\implies a_1=-2,a_0=-1,b=3$

So you have

$\displaystyle\int_0^2\frac{x^2+x+1}{(x+1)^2(x+2)}\,\mathrm dx=-2\int_0^2\frac x{(x+1)^2}\,\mathrm dx-\int_0^2\frac{\mathrm dx}{(x+1)^2}+3\int_0^2\frac{\mathrm dx}{x+2}$
$=\displaystyle-2\int_1^3\dfrac{x-1}{x^2}\,\mathrm dx-\int_0^2\frac{\mathrm dx}{(x+1)^2}+3\int_0^2\frac{\mathrm dx}{x+2}$

where in the first integral we substitute $x\mapsto x+1$ .

$=\displaystyle-2\int_1^3\left(\frac1x-\frac1{x^2}\right)\,\mathrm dx-\frac1{1+x}\bigg|_{x=0}^{x=2}+3\ln|x+2|\bigg|_{x=0}^{x=2}$
$=-2\left(\ln|x|+\dfrac1x\right)\bigg|_{x=1}^{x=3}-\dfrac23+3(\ln4-\ln2)$
$=-2\left(\ln3+\dfrac13-1\right)-\dfrac23+3\ln2$
$=\dfrac23+\ln\dfrac89$