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Rewriting before integrating

AP.CALC:
FUN‑6 (EU)
,
FUN‑6.C (LO)
,
FUN‑6.C.1 (EK)
,
FUN‑6.C.2 (EK)
Some indefinite integrals are much simpler to integrate by algebraically rewriting the integrand first.

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• At , I attempted to rewrite it a different way, which ended up being wrong. Why is it wrong?
I went::
(x^3 + 3x^2 - 5) / x^2
Then I converted the denominator to x^-2, and multiplied it in front of the numerator. Isn't it the same as dividing it?
This resulted: x^-2 (x^3 + 3x^2 - 5)
I multiplied everything inside the brackets with the x to the negative to power and got:
x^-1 + 3x^0 - 5

So integrating it became:
x^0+ 3x - 5x + c

Yikes....
• It's wrong because you didn't multiply everything inside the brackets correctly.
x¯² ( x³ + 3x² - 5) = x + 3 - 5x¯²
Integrating would give x²/2 + 3x + 5/x + C
• At Sal rewrites the equation by dividing every term solely for x^2. However, this new equation would have a new domain that does include x=0. If you were to use defined integral for integration over some range including 0, wouldn't this integral be incorrect?
• The domain of the new equation doesn't include 0, since 0^(-2) is undefined. But even if it did, this wouldn't affect any definite integral we take, since the area above the point 0 contributes no area (it's a rectangle with width 0).
• I'm aware that the derivative of an expression + c is equal to just the expression, but why/how doesn't that change the actual area under the curve?
• Changing the value of +C does change the area under the curve. What it doesn't change is the derivative; shifting the function up and down doesn't affect the slope of the function at a given x-coordinate.
• For the first example, is there such a thing as a "product rule" for integration?
(1 vote)
• ok just curious. On the first problem I got ((3x^4)/3)-((x^3)/3) + 2C. I got the 2C from adding F(3x^3) and F(-x^2), each of course coming with their own C.

I get that the C is a random constant so 2C and C are practically the same thing, but does it ever matter?
• What would you do if you had two functions being divided or multiplied and there was no easy way to simplify the function algebraically? What would you do next, some type of reverse rule?
• At Sal mentions the reverse power rule. Can someone explain it?
(1 vote)
• How can you take an antiderivative when the exponent is -1? (like x^-1)
(1 vote)
• Good question! Note that we can’t use the reversed power rule because we would end up dividing by zero.

Instead, we note that the derivative of ln|x| is 1/x, which is x^(-1).

Therefore, since the derivative of a constant term is zero, the most general antiderivative of x^(-1) is ln|x| + c.

Have a blessed, wonderful day!