Show Wolfram Alpha's answer is equivalent to my answer.

[jlmccart03]

Homework Statement

Integrate x²(2+x³)⁴dx.
Show that Wolfram Alpha's answer is equivalent to your answer.

Homework Equations

No equations besides knowing that the integral of x^power is 1/power+1 * x^{power + 1}

The Attempt at a Solution

So I have the answer to the integral by hand as (2+x³)⁵)/15 + C.
When I go to Wolfram Alpha it gives x¹⁵/15 + 2x¹²/3 + 8x⁹/3 + 16x⁶/3 + 16x³/3 + C

I really truly have no idea how these two are the same. I tried multiple types of manipulation to my answer, but I am completely lost on where basically every factor comes from besides the first x¹⁵/15.

Any help will be appreciated!

 

[member 587159]

Your answer is correct.

They expanded $(2+x^3)^5$ as $x^{15} + 10 x^{12} + 40 x^9 + 80 x^6 + 80 x^3 + 32$ (this can be found using Newton's binomium). Thus, after dividing both sides with 15, we get:

$\frac{(2+x^3)^5}{15} = x^{15} /15 + 2x^{12}/3 + 8x^9/3 + 16x^6/3 + 16x^3/3 + 32/15$

However, the constant in the primitive function does not matter (as we write $+ c$ anyway), so we can drop the $32/15$ safely.

 

[jlmccart03]

Your answer is correct.

They expanded $(2+x^3)^5$ as $x^{15} + 10 x^{12} + 40 x^9 + 80 x^6 + 80 x^3 + 32$ (this can be found using Newton's binomium). Thus, after dividing both sides with 15, we get:

$\frac{(2+x^3)^5}{15} = x^{15} /15 + 2x^{12}/3 + 8x^9/3 + 16x^6/3 + 16x^3/3 + 32/15$

However, the constant in the solution of an indefinite integral does not matter, so we can drop the $32/15$ safely.

Ok, So they simply expanded the numerator using a thing called Newton's binomium? I will do some research on that, but I do not think we have ever learned what Newton's Binomium is. Thanks for explaining how this worked. I was completely lost on how it worked, but now it seemed relatively simple.

 

[member 587159]

You could just have done the multiplications by hand using the distribiutivity property, but the Binomial theorem would be faster:

https://en.wikipedia.org/wiki/Binomial_theorem

 

[jlmccart03]

You could just have done the multiplications by hand using the distribiutivity property, but the Binomial theorem would be faster:

https://en.wikipedia.org/wiki/Binomial_theorem

OHHHHHHH that is what that is called. Ok, so I have done that before. Totally did not think of that as a solution. Thanks for the link, totally forgot that I could use that method.