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

In summary, Wolfram Alpha's answer is equivalent to my answer as it provides a comprehensive and accurate solution to the given problem or question. It utilizes advanced algorithms and data sources to generate a detailed and reliable output that can be used for various purposes such as education, research, and data analysis. Its user-friendly interface and wide range of features make it a valuable tool for obtaining quick and precise answers. Overall, Wolfram Alpha's answer can be trusted and is a valuable resource for those seeking accurate information.
  • #1
jlmccart03
175
9

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 xpower is 1/power+1 * xpower + 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!
 
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  • #2
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.
 
  • #3
Math_QED said:
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.
 
  • #5
Math_QED said:
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.
 

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