Integral of 1 / (x^2 + 2) dx ?

[mitochan]

Is there some simple easy trick I am missing that makes this much easier than ? You can tell me privately if you wish.

Please take a look at attached calculation.

 

[bob012345]

Please take a look at attached calculation.

Thanks. It does seem simpler with the $1$ in the denominator.

 

[Delta2]

@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 $A$.

But for x=3i makes some of the denominators of the identity zero...

 

[bob012345]

But for x=3i makes some of the denominators of the identity zero...

If you multiply both sides by one of the factors, say $x-3i$, solve for $A$ and then set $x=3i$ it works. Cancel those terms before you make the substitution.

 

[Tapias5000]

But for x=3i makes some of the denominators of the identity zero...

That's the idea, so you can find A, then do x=-3i and find B.

 

[bob012345]

That's the idea, so you can find A, then do x=-3i and find B.

I think it is being suggested this is like the proof $1=2$.

 

[Tapias5000]

If you multiply both sides by one of the factors, say $x-3i$, solve for $A$ and then set $x=3i$ it works. Cancel those terms before you make the substitution.

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.

 

[Delta2]

It should not have been left to the reader to do first the multiplication of the identity by x-3i, it should have been explicitly mentioned by @ergospherical :D.

 

[vela]

But for x=3i makes some of the denominators of the identity zero...

You can think of it like this:
\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.

 

[Mark44]

Please take a look at attached calculation.

This is exactly how I would do it.

 

[Mark44]

I think it is being suggested this is like the proof $1=2$.

No, not at all. I think you might be referring to the post by ergospherical.

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 $A$.

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:
$1 = A(x - 3i) + B(x + 3i)$
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.

 

[Tapias5000]

It should not have been left to the reader to do first the multiplication of the identity by x-3i, it should have been explicitly mentioned by @ergospherical :D.

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

 

[mitochan]

Thanks. It does seem simpler with the $1$ in the denominator.

Then
$$\int \frac{dx}{x^2+2}= \frac{1}{\sqrt{2}}\int\frac{d\frac{x}{\sqrt{2}}}{(\frac{x}{\sqrt{2}})^2+1}=...$$

 

[bob012345]

Then
$$\int \frac{dx}{x^2+2}= \frac{1}{\sqrt{2}}\int\frac{d\frac{x}{\sqrt{2}}}{(\frac{x}{\sqrt{2}})^2+1}=...$$

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

 

[Mark44]

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

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.

 

[bob012345]

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.

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.

 

[vela]

Now try it with a hyperbolic trig substitution.

 

[bob012345]

Now try it with a hyperbolic trig substitution.

I tried it and it was not easier by any means.

 

[plum356]

I think that the most straightforward way is to factorise $x^2+2$ into $2\left(1+x^2/2\right)=2\left[1+(x/\sqrt 2)^2\right]$.
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.

 

[benorin]

You guys have beaten this integral in a few ways, OP did you follow any of them? The easy way involves $1+\left(\tfrac{x}{\sqrt{2}}\right) ^2$ and a trig substitution. You probably haven’t learned series methods yet but that would of course be much harder.

I think you can safely lock this thread.