Simple indefinite integral question

Questions
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1

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?

2

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
3

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

4

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”.