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

[Petrus]

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

Regards,

 

[Ackbach]

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

 

[Petrus]

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.

 

[Ackbach]

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?

 

[Petrus]

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

 

[alyafey22]

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

 

[Ackbach]

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.

 

[Petrus]

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?

 

[Ackbach]

$$8x-\frac{16}{\sqrt{2}}\arctan(\frac{x}{\sqrt{2}})+C$$ 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.

 

[Petrus]

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]

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]

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]

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].

 

[Ackbach]

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.