Question #8e17c

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Question #8e17c

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Answer 1 · 2017-02-27T01:05:01

$ln ∣ ∣ ∣ \frac{x - 2}{x - 3} ∣ ∣ ∣ + ln | x - 2 |$ $ln|(x - 2)/(x - 3)| + ln|x-2|$

Explanation:

Right off the bat, we can see that we will need to use impartial fraction integrals.

We set it as $\int \frac{a}{x - 3} d x + \frac{b}{x - 2} d x$ $inta/(x-3)dx + b/(x-2)dx$
$a x + b x = 1$ $ax + bx = 1$
$a + b = 1$ $a + b = 1$
$- 2 a - 3 b = - 4$ $-2a - 3b = -4$

Solving it out, we get $a = - 1$ $a = -1$ and $b = 2$ $b = 2$

$\int \frac{- 1}{x - 3} d x + \int \frac{2}{x - 2} d x$ $int(-1)/(x-3)dx+int2/(x-2)dx$
$- \int \frac{1}{x - 3} d x + 2 \int \frac{1}{x - 2} d x$ $-int1/(x-3)dx+2int1/(x-2)dx$
Integral this $- \int \frac{1}{x - 3} d x + \int \frac{1}{x - 2} d x + \int \frac{1}{x - 2} d x$ $-int1/(x-3)dx+int1/(x-2)dx+int1/(x-2)dx$

to get this $- ln | x - 3 | + 2 ln | x - 2 |$ $-ln|x-3| + 2ln|x-2|$

Now to integral ${[- ln | x - 3 | + 2 ln | x - 2 |]}_{0}^{1}$ $[-ln|x-3| + 2ln|x-2|]_0^1$

$(- ln | 1 - 3 | + 2 ln | 1 - 2 |) - (- ln | 0 - 3 | + 2 ln | 0 - 2 |)$ $(-ln|1-3| + 2ln|1-2|) - (-ln|0-3| + 2ln|0-2|)$
= $(- ln | - 2 | + 2 ln | - 1 |) - (- ln | - 3 | + 2 ln | - 2 |)$ $(-ln|-2| + 2ln|-1|) - (-ln|-3| + 2ln|-2|)$
= $(- ln 2 + ln 1) - (- ln 3 + 2 ln 2)$ $(-ln2 + ln1) - (-ln3 + 2ln2)$
= $- ln 2 - \frac{ln 4}{3}$ $-ln2 - ln4/3$
= $- \frac{ln 8}{3}$ $-ln8/3$
= $- .981$ $-.981$

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Question #8e17c

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