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

In summary: 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...?
Mentor note: Moved from technical section, so missing the homework template.
How do you integrate this?

$$\int \frac{1}{x^2 + 2} dx$$

My attempt is $$\ln |x^2 + 2| + C$$

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Did you try differentiating your answer to see whether you get the integrand?

How do you integrate this?

$$\int \frac{1}{x^2 + 2} dx$$

My attempt is $$\ln |x^2 + 2| + C$$
At least this time you showed an attempt. You can always check your work with an indefinite integral by differentiating your answer. If you do this and get the original integrand, your work is correct.

Note that ##\ln|x^2 + 2| = \ln(x^2 + 2)##, so the absolute value isn't needed.

$$\frac d {dx} \left( \ln(x^2 + 2)\right) = \frac 1 {x^2 + 2} \cdot \frac d {dx} (x^2 + 2)$$
Does that work out to your integrand?

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<Post edited by a mentor, to place more of the burden on the thread starter.>
##
\begin{array}{l}\int _{ }^{ }\frac{2x}{x^2+2}dx\ \left(u=x^2+2,\ dx=\frac{du}{2x}\right)\\

\ln \left|x^2+2\right|+C\\\end{array}
##
obviously modifying the exercise is not valid.
Instead what you have to do is a trigonometric substitution.

Last edited by a moderator:
Tapias5000 said:
<Post edited by a mentor, to place more of the burden on the thread starter.>
##
\begin{array}{l}\int _{ }^{ }\frac{2x}{x^2+2}dx\ \left(u=x^2+2,\ dx=\frac{du}{2x}\right)\\

\ln \left|x^2+2\right|+C\\\end{array}
##
obviously modifying the exercise is not valid.
Instead what you have to do is a trigonometric substitution.

This is what I am looking for. Before my question is deleted, I asked for clues and hints. But my question is deleted because I don't write my attempt. I thought trigonometric substitution only work if there is available a square root such as ##\sqrt{x^2 + 2}##.

This is what I am looking for. Before my question is deleted, I asked for clues and hints. But my question is deleted because I don't write my attempt. I thought trigonometric substitution only work if there is available a square root such as ##\sqrt{x^2 + 2}##.
You don't have to have a root to do it, you could even do a trigonometric substitution in this integral
## \int _{ }^{ }\left(x^2+2\right)dx ## and would get the same response as doing it the quickest way.

PeroK
.answer is of the form ##Aarctan(Bx)+C##. Work out the constants

If you can integrate
$$\int \frac{1}{x^2+1} dx$$
you are not far from the answer.

Delta2
You don't need a trigonometric sub for this one, a substitution of the form ##y=Ax## (what the constant A should be equal to?) and the hint from the previous post is enough to solve the problem.

I thought trigonometric substitution only work if there is available a square root such as ##\sqrt{x^2 + 2}##.
No. Trig substitution is helpful if you have the sum or difference of squared terms - they don't need to be inside a square root.

Delta2
Sorry, I didn't mean to give too much help. Is this an actual homework problem or just a question? From the format it is not clear to me. Thanks.

The thread probably should have been in the homework forum.

bob012345
vela said:
The thread probably should have been in the homework forum.
Belatedly moved...

SammyS and bob012345
Why not let ##u = x/\sqrt{2}##?

Delta2
mitochan said:
If you can integrate
$$\int \frac{1}{x^2+1} dx$$
you are not far from the answer.
Is there some simple easy trick I am missing that makes this much easier than ##\int \frac{1}{x^2+2} dx##? You can tell me privately if you wish.

bob012345 said:
Is there some simple easy trick I am missing that makes this much easier than ##\int \frac{1}{x^2+2} dx##? You can tell me privately if you wish.
Do you watch blackpenredpen ever? His strategy to integrals is sometimes "wouldn't it be nice if we could take [annoyingly difficult integral] and convert it into [something easy] to integrate. A simple, but hard to spot substitution can actually convert that integral into something that looks A LOT like the quoted integral.

Also the integral of 1/x^2+1 is a standard integral, evaluating to arctan(x) + C.

Mayhem said:
Do you watch blackpenredpen ever? His strategy to integrals is sometimes "wouldn't it be nice if we could take [annoyingly difficult integral] and convert it into [something easy] to integrate. A simple, but hard to spot substitution can actually convert that integral into something that looks A LOT like the quoted integral.

Also the integral of 1/x^2+1 is a standard integral, evaluating to arctan(x) + C.
I know you can make it look like ##\int \frac{dx}{x^2+1}##with a trivial substitution which was already given in this thread but the question is then is there a clever trick that makes it much easier to integrate than ##\int \frac{dx}{x^2+2}##using other substitutions after that?

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 not?

bob012345 said:
I know you can make it look like ##\int \frac{dx}{x^2+1}##with a trivial substitution which was already given in this thread but the question is then is there a clever trick that makes it much easier to integrate than ##\int \frac{dx}{x^2+2}##using other substitutions after that?

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 not?
Why would you need a trig substitution? Once you do the clever substitution, the answer practically gives itself.

I have no idea how to arrive at arctan(x) by integrating that. But it is actually fairly easy to prove that (arctan(x))' = 1/x2+2. Thus, using F'(x)=f(x), you can argue your case easily. Maybe you can find a trick to rewrite the integral such that you can do a partial fraction decomposition, but that just seems like a lot of work.

Mayhem said:
Why would you need a trig substitution? Once you do the clever substitution, the answer practically gives itself.

I have no idea how to arrive at arctan(x) by integrating that. But it is actually fairly easy to prove that (arctan(x))' = 1/x2+2. Thus, using F'(x)=f(x), you can argue your case easily. Maybe you can find a trick to rewrite the integral such that you can do a partial fraction decomposition, but that just seems like a lot of work.
My table of Integrals has the general case ##\int \frac{dx}{x^2+a^2}## as a 'standard' integral too, not just the case where ##a=1##. But I think this being a Calculus homework problem it is expected of the OP to figure out how to do it. Hints were given above.

Mayhem said:
Why would you need a trig substitution? Once you do the clever substitution, the answer practically gives itself.
I'm intrigued. If the answer is an inverse trig function, then how can trig functions be avoided?

PeroK said:
I'm intrigued. If the answer is an inverse trig function, then how can trig functions be avoided?
Perhaps the confusion lies in me not evaluating 1/x^2+1 manually but simply using the standard integral. Thinking about it, that integral could probably be solved with trig substitution.

Mayhem said:
Why would you need a trig substitution?
See below.
Mayhem said:
I have no idea how to arrive at arctan(x) by integrating that.
It can be done very easily by using a trig substitution; namely ##x = \tan(\theta)##. With this substitution, the integration becomes trivial.
Mayhem said:
But it is actually fairly easy to prove that (arctan(x))' = 1/x2+2
You need more parentheses. Taken literally, what you have written (twice) would be interpreted to mean this:
$$\frac 1 {x^2} + 2$$

Without LaTeX, the above should be written as 1/(x^2 + 2).

SammyS
PeroK said:
I'm intrigued. If the answer is an inverse trig function, then how can trig functions be avoided?
You could use imaginary numbers then partial fractions and you will get 2 logarithms, so you can skip the trigonometric substitution in this case.

Mayhem and PeroK
Tapias5000 said:
You could use imaginary numbers then partial fractions and you will get 2 logarithms, so you can skip the trigonometric substitution in this case.
How about an unrelated example so as not to directly solve this integral but to illustrate the principle?

bob012345 said:
How about an unrelated example so as not to directly solve this integral but to illustrate the principle?
The main part is the algebra involved. Suppose we have ##\frac 1 {x^2 + 9}##.

$$\frac 1 {x^2 + 9} = \frac 1 {(x + 3i)(x - 3i)} = \frac A {x + 3i} + \frac B {x - 3i}$$
After finding the constants A and B it's relatively simple to compute the integral ##\int \frac {dx}{x^2 + 9}## as it can now be written as two simpler integrals.

bob012345 and Tapias5000
Tapias5000 said:
You could use imaginary numbers then partial fractions and you will get 2 logarithms, so you can skip the trigonometric substitution in this case.
And it's left as an exercise to the reader to show that one solution can be transformed into the other.

Tapias5000 and PeroK
bob012345 said:
How about an unrelated example so as not to directly solve this integral but to illustrate the principle?
That's the idea, it also works when x^2 is being multiplied by a number like for example:

## \int _{ }^{ }\frac{1}{5x^2+9}dx ##
## 5x^2+9=0 ##
##x^2=-\frac{9}{5} ##
## x=\frac{3i}{\sqrt{5}},\ \ x=-\frac{3i}{\sqrt{5}}\ →\ \left(\sqrt{5}x-3i\right)\left(\sqrt{5}x+3i\right)##
##\frac{1}{\left(\sqrt{5}x-3i\right)\left(\sqrt{5}x+3i\right)}=\frac{A}{\left(\sqrt{5}x-3i\right)}+\frac{B}{\left(\sqrt{5}x+3i\right)} ##

Basically we can generalize this type of integral to:

## \int _{ }^{ }\frac{1}{nx^2+a}dx ## where n≠0 and a≠0 where a and n are a number with a positive sign
obtaining ##\int _{ }^{ }\frac{1}{\left(\sqrt{n}x-\sqrt{a}i\right)\left(\sqrt{n}x+\sqrt{a}i\right)}dx ##

bob012345
Mark44 said:
The main part is the algebra involved. Suppose we have ##\frac 1 {x^2 + 9}##.

$$\frac 1 {x^2 + 9} = \frac 1 {(x + 3i)(x - 3i)} = \frac A {x + 3i} + \frac B {x - 3i}$$
After finding the constants A and B it's relatively simple to compute the integral ##\int \frac {dx}{x^2 + 9}## as it can now be written as two simpler integrals.
I assume the integration is just over real numbers and the ##i## is treated as a constant? Don't answer I'll figure that out...but frankly, this seems to be a lot harder way to do it.

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Mark44 said:
The main part is the algebra involved. Suppose we have ##\frac 1 {x^2 + 9}##.

$$\frac 1 {x^2 + 9} = \frac 1 {(x + 3i)(x - 3i)} = \frac A {x + 3i} + \frac B {x - 3i}$$
After finding the constants A and B it's relatively simple to compute the integral ##\int \frac {dx}{x^2 + 9}## as it can now be written as two simpler integrals.
I tried but I cannot see how ##A,B## are constants?

@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##.

ergospherical said:
@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##.
I originally assumed ##A,B## were real leading to trouble. That's when I made that comment. I realized later they might be complex but it was too late. Thanks!

bob012345 said:
I tried but I cannot see how ##A,B## are constants?
Both are imaginary constants. I get A = i/6 and B = -i/6.

bob012345
Mark44 said:
Both are imaginary constants. I get A = i/6 and B = -i/6.
I got that too. Now for fun I'll try the original integral this way but quietly...

In my class I taught this as a trig substitution. I.e. we know (at least) two basic trig identities with squares in them: sin^2+cos^2=1, and 1+tan^2 = sec^2, (obtained by dividing the first one by cos^2). the first one gives cos^2 = 1-sin^2, and so we have identities that can be used to simplify 1-x^2 as well as 1+x^2, by putting x = sin(t) or x = tan(t).

To integrate dx/(1+x^2), we put x = tan(t), and dx = sec^2(t)dt, and get the integral as t. and since x= tan(t), we have t = arctan(x).

in your case you have 2+x^2 which looks like 1+tan^2, except off by a constant. now constant multipliers do no harm so you could try setting 2+x^2 = c(1+tan^2(t)), and go from there.

complex methods are also fun and illuminating, but sometimes may give non real answers. of course if you know e^it = cos(t) + i sin (t), you can often find your way back.

In fact when you look at complex numbers and complex path integrals, log and arctan are somewhat the same, except for interchanging i and -i with 0 and infinity, since integration of 1/(1+z^2) behaves the same as you go around i and -i as integrating 1/z does as you go around 0 and infinity!

bob012345 and PeroK
bob012345 said:
I got that too. Now for fun I'll try the original integral this way but quietly...
Finished and got the same results but converting from the natural log form was not trivial unless one already knew about the relationship as mentioned by @mathwonk above, which I didn't but now I do. So, yes, it can be done without a trig substitution but to me, not as easily.

But in a way both are doing similar things in that one is mapping the problem from a one dimensional space to a two dimensional space, the ##x,y## plane with an angle in one case and the complex plane in the other.

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