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Antiderivative

Petrus

Well-known member
Feb 21, 2013
739
Hello MHB,
I got problem to antiderivate \(\displaystyle \frac{8x^2}{x^2+2}\)

Regards,
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
I would probably go for polynomial long division here. What do you get?
 

Petrus

Well-known member
Feb 21, 2013
739
I would probably go for polynomial long division here. What do you get?
I get \(\displaystyle 8- \frac{16}{x^2+2}\) it should be correct.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
I get \(\displaystyle 8- \frac{16}{x^2+2}\) 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?
 

Petrus

Well-known member
Feb 21, 2013
739
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 .. \(\displaystyle 8x-\frac{8}{x}\ln(x^2+2)\) and that is wrong
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
One simple technique that you can use is adding and subtracting

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

Now separate to get

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

Now , you should use \(\displaystyle \tan ^{-1}\) to integrate it
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
I only get .. \(\displaystyle \frac{8}{x}\ln(x^2+2)\) 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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Petrus

Well-known member
Feb 21, 2013
739
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?
\(\displaystyle 8x-\frac{16}{\sqrt{2}}\arctan(\frac{x}{\sqrt{2}})-16C\) is that correct?
 
Last edited by a moderator:

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
\(\displaystyle 8x-\frac{16}{\sqrt{2}}\arctan(\frac{x}{\sqrt{2}})+C\) 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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Petrus

Well-known member
Feb 21, 2013
739
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,
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
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".
 

Petrus

Well-known member
Feb 21, 2013
739
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,
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
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 \(\displaystyle \displaystyle \int{8 - \frac{16}{x^2 + 2}\,dx} \).
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
No, the integral was actually \(\displaystyle \displaystyle \int{8 - \frac{16}{x^2 + 2}\,dx} \).
You're quite right. Thanks for the catch! I'll edit previous posts to reflect that fact.