`int 8x^6 dx`

`8` is a constant, so it can go out the front:

`8 int x^6 dx`

Next, do the integration step by adding 1 to the index and dividing by the new number:

`8 x^(6+1)/(6+1) = 8(x^7)/7`

And of course, we must not forget the constant.

So the final answer is:

`int 8x^6 dx = 8(x^7)/7+K`

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