# Anti derivative

• Jun 22nd 2008, 08:59 AM
PensFan10
Anti derivative
I have to find the anti derivative of x(x^2+2)^2
This is what I did:

f ' (x) = ( x^2 + 2)^2

= X (X^4 + 4x^2 + 4)

F(x) = (1/2x^2) (1/5x^5 + 4/3x^3 + C)

The book did not show any examples of problems like this and I was not sure if there was a product rule that I needed or if the Chain Rule came into play somehow.

Thanks.
• Jun 22nd 2008, 09:03 AM
Mathstud28
Quote:

Originally Posted by PensFan10
I have to find the anti derivative of x(x^2+2)^2
This is what I did:

f ' (x) = ( x^2 + 2)^2

= X (X^4 + 4x^2 + 4)

F(x) = (1/2x^2) (1/5x^5 + 4/3x^3 + C)

The book did not show any examples of problems like this and I was not sure if there was a product rule that I needed or if the Chain Rule came into play somehow.

Thanks.

$\displaystyle \int{x(x^2+2)^2{dx}}$

Way one

$\displaystyle x(x^2+2)^2=x(x^4+4x^2+4)=x^5+4x^3+4x$

From there I am sure you can integrate

Or you can do this

Let $\displaystyle u=x^2+2$

so $\displaystyle du=2x\text{ }dx$

So we have

$\displaystyle \frac{1}{2}\int{u^2}du=\frac{u^3}{6}+C$

Now back sub.
• Jun 22nd 2008, 09:06 AM
PensFan10
okay cool
Okay so the mistake I made was that I should have distributed the x. I appreciate the speedy and helpful response.
• Jun 22nd 2008, 09:07 AM
Moo
Hello :)

Quote:

Originally Posted by PensFan10
I have to find the anti derivative of x(x^2+2)^2
This is what I did:

f ' (x) = x( x^2 + 2)^2

= X (X^4 + 4x^2 + 4)

F(x) = (1/2x^2) (1/5x^5 + 4/3x^3 + C) << SiMoon says : the antiderivative of a product is not the product of the antiderivatives ! If you wanted to do it this way, expan x(x^4+4x^2+4)=x^5+4x^3+4x and then integrate the sum.

The book did not show any examples of problems like this and I was not sure if there was a product rule that I needed or if the Chain Rule came into play somehow.

Thanks.

Let $\displaystyle f(x)=x^2+2$.

You can notice that $\displaystyle f'(x)=2x$.

So $\displaystyle x=\frac{f'(x)}{2}$

You have to find the antiderivative of $\displaystyle x(x^2+2)=\frac 12 \cdot f'(x) \cdot (f(x))^2$

Note that the derivative of $\displaystyle [f(x)]^n=n \cdot f'(x) \cdot [f(x)]^{n-1}$

Here, n-1=2. So n=3.

--->< $\displaystyle \frac 12 \cdot f'(x) \cdot [f(x)]^2$ is the derivative of $\displaystyle \frac 12 \cdot \frac 13 \cdot [f(x)]^3$

Antiderivative of x(x^2+2)^2 is $\displaystyle \boxed{\frac 16 \cdot (x^2+2)^3+C}$

Edit : a bit slow today... Attachment 6917
• Jun 22nd 2008, 09:22 AM
PensFan10
Question
This is probably a silly question, but will your answer give the same as mine which is 1/6x^6+x^4+2x^2 + C

The reason I ask this question is because I believe (i am not 100% sure) that your answer would give a number. what i mean by this is that when you expand your cube, wont a factor of 2 be in the answer.
I know i didnt do the best job of explaining what I mean.

Also, could you tell me the formula for expanding a cube.

Thanks
• Jun 22nd 2008, 10:02 AM
Moo
Quote:

Originally Posted by PensFan10
This is probably a silly question, but will your answer give the same as mine which is 1/6x^6+x^4+2x^2 + C

The reason I ask this question is because I believe (i am not 100% sure) that your answer would give a number. what i mean by this is that when you expand your cube, wont a factor of 2 be in the answer.
I know i didnt do the best job of explaining what I mean.

Also, could you tell me the formula for expanding a cube.

Thanks

$\displaystyle (a+b)^3=a^3+3ab^2+3a^2b+b^3$

It won't give a number :eek:
You will just be able to remove $\displaystyle 2^3$ in the expansion because C is a constant and so is $\displaystyle 2^3$ (Tongueout)
• Jun 22nd 2008, 10:10 AM
PensFan10
Okay
The good thing is, it makes sense :D Thanks