- #36

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Please take a look at attached calculation.Is there some simple easy trick I am missing that makes thismucheasier than ? You can tell me privately if you wish.

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I think the issue is for the OP to do the integration and not just look it up as a 'standard' integral. My point is if you do the integration on both, they are the same difficulty are they...?f

- #36

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Please take a look at attached calculation.Is there some simple easy trick I am missing that makes thismucheasier than ? You can tell me privately if you wish.

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- #37

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Thanks. It does seem simpler with the ##1## in the denominator.Please take a look at attached calculation.

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- #38

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But for x=3i makes some of the denominators of the identity zero...@bob012345, multiply the identity ##\dfrac{1}{(x+3i)(x-3i)} = \dfrac{A}{x + 3i} + \dfrac{B}{x-3i}## by the factor ##(x-3i)##, and then set ##x = 3i## (you can do this because the identity holds for all ##x##). And do a similar thing to find ##A##.

- #39

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If you multiply both sides by one of the factors, say ##x-3i##, solve for ##A## andBut for x=3i makes some of the denominators of the identity zero...

- #40

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That's the idea, so you can find A, then do x=-3i and find B.But for x=3i makes some of the denominators of the identity zero...

- #41

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I think it is being suggested this is like the proof ##1=2##.That's the idea, so you can find A, then do x=-3i and find B.

- #42

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Emmm, by multiplying both sides by (x-3i) and then making x=3i, you will not fall into any error, it is valid to do so.If you multiply both sides by one of the factors, say ##x-3i##, solve for ##A## andthenset ##x=3i## it works. Cancel those terms before you make the substitution.

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- #44

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You can think of it like this:But for x=3i makes some of the denominators of the identity zero...

\begin{align*}

\lim_{x \to 3i} (x-3i)\frac{1}{(x-3i)(x+3i)} &= \lim_{x \to 3i} (x-3i)\left[\frac{A}{x-3i}+\frac{B}{x+3i}\right] \\

\lim_{x \to 3i} \frac{1}{x+3i} &= \lim_{x \to 3i} \left[A+B\frac{x-3i}{x+3i}\right] \\

\end{align*} After canceling the factors of ##x-3i##, the functions are continuous at ##x=3i##, so you can evaluate the limit by simply plugging in the value.

- #45

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This is exactly how I would do it.Please take a look at attached calculation.

- #46

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No, not at all. I think you might be referring to the post by ergospherical.I think it is being suggested this is like the proof ##1=2##.

An important point is that the equation above is an identity: one that is is true for all values of the variable x, excepting only ##x = \pm 3i##. For any values of x other than these two, we can multiply both sides of the equation by ##(x + 3i)(x - 3i)## to get this equation:ergospherical said:multiply the identity ##\dfrac{1}{(x+3i)(x-3i)} = \dfrac{A}{x + 3i} + \dfrac{B}{x-3i}## by the factor ##(x-3i)##, and then set ##x = 3i## (you can do this because the identity holds for all ##x##). And do a similar thing to find ##A##.

##1 = A(x - 3i) + B(x + 3i)##

The new equation is still an identity: it must be true for any value of x. The way I chose to go is to solve for A and B like so:

##1 = (A + B)x - 3i(A - B)

For this to be true for all values of x, A + B must be 0 (there is no x term on the left side), and -3i(A - B) = 1.

Solve these two equations for A and B, and Bob's your uncle.

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- #47

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hehe, I use Quora more so my answers are usually complete aum, I'm not used to this forum at all hehe.

- #48

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ThenThanks. It does seem simpler with the ##1## in the denominator.

[tex]\int \frac{dx}{x^2+2}= \frac{1}{\sqrt{2}}\int\frac{d\frac{x}{\sqrt{2}}}{(\frac{x}{\sqrt{2}})^2+1}=...[/tex]

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- #49

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I just used a basic trig with ##tan(\theta)= \large\frac{x}{a}## and it worked fine.Then

[tex]\int \frac{dx}{x^2+2}= \frac{1}{\sqrt{2}}\int\frac{d\frac{x}{\sqrt{2}}}{(\frac{x}{\sqrt{2}})^2+1}=...[/tex]

- #50

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The image in post #36 shows why this substitution works. For ##x^2 + 2##, the length of the leg adjacent to the angle is ##\sqrt 2## rather than 1.I just used a basic trig with ##tan(\theta)= \large\frac{x}{a}## and it worked fine.

- #51

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Of course and the point I made was the method I used works for any ##a##. That's why I used ##a## and not ##1##. I found it easier than manipulating the integral first to make it a ##1## or using complex numbers.The image in post #36 shows why this substitution works. For ##x^2 + 2##, the length of the leg adjacent to the angle is ##\sqrt 2## rather than 1.

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- #52

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Now try it with a hyperbolic trig substitution.

- #53

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I tried it and it was not easier by any means.Now try it with a hyperbolic trig substitution.

- #54

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When you see an integral close to one that you would usually find in an integration table, ##(1+x^2)^{-1}## for instance, try adding a ##0## or factorising something.

- #55

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I think you can safely lock this thread.

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