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Petrus
Well-known member
- Feb 21, 2013
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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,
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?
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.
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?
You have to integrate term-by-term now. You haveI only get .. \(\displaystyle \frac{8}{x}\ln(x^2+2)\) and that is wrong
\(\displaystyle 8x-\frac{16}{\sqrt{2}}\arctan(\frac{x}{\sqrt{2}})-16C\) 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?
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?
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.
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,
Hello Ackbach,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".
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?
You're quite right. Thanks for the catch! I'll edit previous posts to reflect that fact.No, the integral was actually \(\displaystyle \displaystyle \int{8 - \frac{16}{x^2 + 2}\,dx} \).