What is the integral of (6x^2 + 2/x^2 + 5) with respect to x?

When we think of integration, we should be thinking of the same method every time; adding one to the power, and then dividing by that new power.The integral of the sum of each of these terms is equal to the sum of the integrals of each individual term, and therefore we can take them on one-by-one.When integrating 6x^2, we use the method i mentioned above, and add one to the power (which is 2 here) and divide by it. So this becomes (6x^3)/3, which equals 2x^3.Now, for the second term, we know that 2/x^2 is the same as 2x^-2. This makes it so much easier to deal with. So with the same method as before, this becomes (2x^-1)/(-1), remembering that when we add 1 to -2, we get -1 and not -3. This then equals -2x^-1, or -2/x.With the last term, 5, there is no power of x. So writing 5 on it's own is technically the same as writing 5x^0, as anything to the power of 0 is 1, and 5*1=5. This makes it easier to visualise what we're trying to do when we integrate this term. So, with the same method as always, this becomes (5x^1)/1, which is just 5x.Now a fairly common mistake when it comes to indefinite integrals is made at this last part. There could always have been a constant value 'C' that was differentiated to give 0 in the term that we're integrating, so we have to add it back on when we've integrated all the other terms, to make sure we cover this possibility.So, adding all the terms back together, the answer is 2x^3 - 2x^-1 + 5x + C.