MHB How do I antiderivate \frac{8x^2}{x^2+2}?

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Hello MHB,
I got problem to antiderivate $$\frac{8x^2}{x^2+2}$$

Regards,
 
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I would probably go for polynomial long division here. What do you get?
 
Ackbach said:
I would probably go for polynomial long division here. What do you get?
I get $$8- \frac{16}{x^2+2}$$ it should be correct.
 
Petrus said:
I get $$8- \frac{16}{x^2+2}$$ it should be correct.

That's what I get, too. And you can check it by simply getting a common denominator and adding back up. So where do you go from here?
 
Ackbach said:
That's what I get, too. And you can check it by simply getting a common denominator and adding back up. So where do you go from here?
I only get .. $$8x-\frac{8}{x}\ln(x^2+2)$$ and that is wrong
 
One simple technique that you can use is adding and subtracting

$$\frac{8x^2+16-16 }{x^2+2}= \frac{8(x^2+2) - 16}{x^2+2}$$

Now separate to get

$$\frac{8(x^2+2) - 16}{x^2+2} = \frac{8(x^2+2)}{x^2+2} -\frac{16}{x^2+2}=8-\frac{16}{x^2+2}$$

Now , you should use $$\tan ^{-1}$$ to integrate it
 
Petrus said:
I only get .. $$\frac{8}{x}\ln(x^2+2)$$ and that is wrong

You have to integrate term-by-term now. You have
$$\int \left( 8- \frac{16}{x^{2}+2} \right)dx
=\int 8 \, dx-16 \int \frac{dx}{x^{2}+2}.$$
Surely you can do the first one. Can you recognize what the second one is?

[EDIT] Prove It saw a mistake in this post, so I have edited out the mistake.
 
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Ackbach said:
You have to integrate term-by-term now. You have
$$\int \left( 8- \frac{16}{x^{2}+2} \right)dx
=\int 8 \, dx-16 \int \frac{dx}{x^{2}+2}.$$
Surely you can do the first one. Can you recognize what the second one is?
$$8x-\frac{16}{\sqrt{2}}\arctan(\frac{x}{\sqrt{2}})-16C$$ is that correct?
 
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Petrus said:
$$8x-\frac{16}{\sqrt{2}}\arctan(\frac{x}{\sqrt{2}})+C$$ is that correct?

You don't have to ask me! Differentiate it and see if you get the original integrand back. By the way (that is, I'm not commenting on the correctness of your answer, merely on its typesetting), I would probably code your answer up as
$$8x-\frac{16}{\sqrt{2}}\tan^{-1} \left(\frac{x}{\sqrt{2}} \right)+C.$$

I prefer $\tan^{-1}$ to $\arctan$, because the notation tells me a little more clearly that this function is the function inverse of $\tan$. Also, I like to use \left and \right to help make the parentheses a better size.
 
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  • #10
Hello.
I get same answer when I derivate so it's correct! This just show that I should train more antiderivate/derivate! Better early then later!:)

Regards,
 
  • #11
Petrus said:
Hello.
I get same answer when I derivate so it's correct! This just show that I should train more antiderivate/derivate! Better early then later!:)

Regards,

The more correct words in English are that you "get the same answer when you differentiate". Better words are "differentiate" and "anti-differentiate" or better yet, "integrate".
 
  • #12
Ackbach said:
The more correct words in English are that you "get the same answer when you differentiate". Better words are "differentiate" and "anti-differentiate" or better yet, "integrate".
Hello Ackbach,
Thanks for correcting me:) In swedish we say 'Derivera' that's why I always say 'Derivate'. I will try remember its called differentiate in english

Regards,
 
  • #13
Ackbach said:
You have to integrate term-by-term now. You have
$$\int \left( x- \frac{16}{x^{2}+2} \right)dx
=\int x \, dx-16 \int \frac{dx}{x^{2}+2}.$$
Surely you can do the first one. Can you recognize what the second one is?

No, the integral was actually [math]\displaystyle \int{8 - \frac{16}{x^2 + 2}\,dx} [/math].
 
  • #14
Prove It said:
No, the integral was actually [math]\displaystyle \int{8 - \frac{16}{x^2 + 2}\,dx} [/math].

You're quite right. Thanks for the catch! I'll edit previous posts to reflect that fact.
 
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