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# Simple indefinite integral question

Posted 10 years ago
 Hello, I am new to Mathematica and I have started using it to help teach myself elementary calculus. I do the problems by hand and then use Mathematica to confirm my answers. This approach worked well with derivatives, but I have noticed that when I use Mathematica to compute simple integrals the output rarely (if ever) matches what I compute by hand with paper and pencil. Even when I use tools like Simplify[] or FullSimplify[] it often doesn't help me. I know there is nothing wrong with the software, but rather something is lacking in my own understanding of what the software is doing or how I am using it. I'm hoping someone can help me understand what I am doing wrong. As an example, Mathematica computes the following simple integral: In[1]:= \[Integral](8 x^2 + 1)^2 16 x \[DifferentialD]x Out[1]= 8 x^2 + 64 x^4 + (512 x^6)/3  Isn't the output missing a term? When I compute the same integral by hand I get an extra term +1/3 so I'm thinking the output should appear as below? Out[1]= 1/3 + 8 x^2 + 64 x^4 + (512 x^6)/3  What am I doing wrong?
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Posted 10 years ago
 Ok, I think I get it. I now see that Mathematica's output was correct. The answer I thought Mathematica should output (including the +1/3 term) would have been incorrect as it only represents one possible antiderivative when there can be infinitely many.Thanks for the help Kay
Posted 10 years ago
 As a matter of fact, the 1/3 term is not an incorrect one, but an incomplete one. Besides the function "Integrate", you can also use "DSolve" to show that constant term which is In:= DSolve[y'[x] == (8 x^2 + 1)^2 16 x, y[x], x] Out = {{y[x] -> 8 x^2 + 64 x^4 + (512 x^6)/3 + C[1]}} The C[1] is just the term missing in the solution by "Integrate".
Posted 10 years ago
 It seems you do an indefinite integral: In[7]:= Integrate[(8 x^2 + 1)^2 16 x, x] Out[7]= 8 x^2 + 64 x^4 + (512 x^6)/3 you can add any constant and differentiation will give you your original equation back In[8]:= D[% + c, x] Out[8]= 16 x + 256 x^3 + 1024 x^5 on the other hand a definite integral from a to be is found like this: In[9]:= Integrate[(8 x^2 + 1)^2 16 x, { x, a, b}] Out[9]= -8 a^2 - 64 a^4 - (512 a^6)/3 + 8 b^2 + 64 b^4 + (512 b^6)/3 
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