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#### Petrus

##### Well-known member

- Feb 21, 2013

- 739

Hello MHB,

I got problem to antiderivate \(\displaystyle \frac{8x^2}{x^2+2}\)

Regards,

I got problem to antiderivate \(\displaystyle \frac{8x^2}{x^2+2}\)

Regards,

- Thread starter Petrus
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- Thread starter
- #1

- Feb 21, 2013

- 739

Hello MHB,

I got problem to antiderivate \(\displaystyle \frac{8x^2}{x^2+2}\)

Regards,

I got problem to antiderivate \(\displaystyle \frac{8x^2}{x^2+2}\)

Regards,

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- Jan 26, 2012

- 4,197

I would probably go for polynomial long division here. What do you get?

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- #3

- Feb 21, 2013

- 739

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

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- #4

- Jan 26, 2012

- 4,197

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 get \(\displaystyle 8- \frac{16}{x^2+2}\) it should be correct.

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- #5

- Feb 21, 2013

- 739

I only get .. \(\displaystyle 8x-\frac{8}{x}\ln(x^2+2)\) and that is wrongThat'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?

- Jan 17, 2013

- 1,667

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

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

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- #7

- Jan 26, 2012

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You have to integrate term-by-term now. You haveI only get .. \(\displaystyle \frac{8}{x}\ln(x^2+2)\) and that is wrong

$$\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.

Last edited:

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- #8

- Feb 21, 2013

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\(\displaystyle 8x-\frac{16}{\sqrt{2}}\arctan(\frac{x}{\sqrt{2}})-16C\) is that correct?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?

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- #9

- Jan 26, 2012

- 4,197

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\(\displaystyle 8x-\frac{16}{\sqrt{2}}\arctan(\frac{x}{\sqrt{2}})+C\) is that correct?

$$8x-\frac{16}{\sqrt{2}}\tan^{-1} \left(\frac{x}{\sqrt{2}} \right)+C.$$

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

- Feb 21, 2013

- 739

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,

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- #11

- Jan 26, 2012

- 4,197

The more correct words in English are that you "get the same answer when you

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,

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- #12

- Feb 21, 2013

- 739

Hello Ackbach,The more correct words in English are that you "get the same answer when youdifferentiate". Better words are "differentiate" and "anti-differentiate" or better yet, "integrate".

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,

No, the integral was actually \(\displaystyle \displaystyle \int{8 - \frac{16}{x^2 + 2}\,dx} \).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?

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- #14

- Jan 26, 2012

- 4,197

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