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

[askor]

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 $$\ln |x^2 + 2| + C$$

 

[PeroK]

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

 

[Mark44]

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?

 

[Tapias5000]

<Post edited by a mentor, to place more of the burden on the thread starter.>
Your answer is incorrect, for your answer to be valid you would have to alter the original problem in this way.
$\begin{array}{l}\int _{ }^{ }\frac{2x}{x^2+2}dx\ \left(u=x^2+2,\ dx=\frac{du}{2x}\right)\\ \ln \left|x^2+2\right|+C\\\end{array}$
obviously modifying the exercise is not valid.
Instead what you have to do is a trigonometric substitution.

 

[askor]

<Post edited by a mentor, to place more of the burden on the thread starter.>
Your answer is incorrect, for your answer to be valid you would have to alter the original problem in this way.
$\begin{array}{l}\int _{ }^{ }\frac{2x}{x^2+2}dx\ \left(u=x^2+2,\ dx=\frac{du}{2x}\right)\\ \ln \left|x^2+2\right|+C\\\end{array}$
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 $\sqrt{x^2 + 2}$.

 

[Tapias5000]

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.

 

[mathman]

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

 

[mitochan]

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.

 

[Mark44]

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.

 

[bob012345]

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.

 

[vela]

The thread probably should have been in the homework forum.

 

[Mark44]

The thread probably should have been in the homework forum.

Belatedly moved...

 

[Mayhem]

Why not let $u = x/\sqrt{2}$?

 

[bob012345]

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.

 

[Mayhem]

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.

 

[bob012345]

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?

 

[Mayhem]

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.

 

[bob012345]

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.

 

[PeroK]

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?

 

[Mayhem]

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.

 

[Mark44]

Why would you need a trig substitution?

See below.

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.

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).

 

[Tapias5000]

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.

 

[bob012345]

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?

 

[Mark44]

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.

 

[vela]

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]

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]

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.

 

[bob012345]

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?

 

[ergospherical]

@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 $A$.