1. Nov 17, 2012

Peppino

I have to find the integral of $$(4-x)x^{-3}$$. My TI-89 says it should be $$\frac{x-2}{x^{2}}+C$$ but I can't seem to get it myself.

I rearranged it to get $$(4x^{-1}-1)x^{-2}$$ and then I used U-Substitution. And set $$U = 4x^{-1}-1$$ so that $$dU = -4x^{-2}dx$$Then I rewrote the integral as $$-\frac{1}{4}\int Udu$$ and evaluated it to get $$-\frac{1}{8}U^{2}+C$$or $$-\frac{1}{8}(4x^{-1}-1)^{2}+C$$ which works out to $$-\frac{(x-4)^{2}}{8x^{2}}+C$$which is not what my calculator says it should be.

What am I doing wrong?

...and if I have to make another Latex equation...

Last edited: Nov 17, 2012
2. Nov 17, 2012

gopher_p

1) If you hadn't dropped the $\frac{1}{8}$, your answer is correct.

3) They're actually the same answer. How can that be?

4) You're making this integral WAY more difficult than it actually is. You don't need a substitution. Why did you just distribute $x^{-1}$ through instead of $x^{-3}$?

3. Nov 17, 2012

Peppino

Right... I meant to put the 1/8 in there. But when I graph the two expressions I get two completely different tables of values. Are you sure they are the same?

As for the unnecessary U-Substitution... that's just me being a moron I guess

4. Nov 17, 2012

gopher_p

What do you see when you look at the graphs? How are the graphs comparable?

What is an indefinite integral? Is it a function, or perhaps something ... broader? Whats that +C for?

5. Nov 17, 2012

Peppino

The table of values are completely different... on my calculator's answer, x = 1 yields -1, while on mine x = 1 yields -1.125.

Is it not a function?

Because an infinite number of different values of C could lead to a function with a derivative of the original function

6. Nov 17, 2012

Peppino

Doing the non-moronic way I get the same answer as my calculator. I'll stick with that for now but any comments will still be appreciated

7. Nov 18, 2012

Staff: Mentor

In other words, carry out the multiplication to arrive at this integral:
$$\int 4x^{-3} - x^{-2}~dx$$

It should be pretty simple after that.

8. Nov 18, 2012

Peppino

Yes, I realized that. Any guess as to why my circuitous method didn't work?

9. Nov 18, 2012

SammyS

Staff Emeritus
$\displaystyle -\frac{(x-4)^{2}}{8x^{2}}-\frac{x-2}{x^2}=-\frac{1}{8}$

The answers differ only by a constant, so both have the same derivative.

10. Nov 18, 2012

Curious3141

To add to what SammyS said, remember that that so-called "arbitrary constant" of indefinite integration is really that - completely arbitrary. Any constant added to your answer will still give a completely valid answer. So when two expressions differ only be a constant value, they are both equally valid solutions.

Even though you wrote the "expected" solution and yours with the constant having the same symbol $C$, remember that they're actually different in value (they differ by $\frac{1}{8}$ in this case). It would be more mathematically correct to write the constant in one case as $C_1$ and the other as $C_2$ so you don't get confused.

11. Nov 18, 2012

Peppino

Oh, I see it. That's interesting. So when using different methods to find an integral, the value of C (or, whats included in C) would change?

12. Nov 18, 2012

SammyS

Staff Emeritus
C can be different.

Two correct answers can differ by a constant.

13. Nov 18, 2012

Staff: Mentor

For a different example that shows how two different techniques can produce what seem to be different antiderivatives, consider
$$\int sin(x)cos(x)dx$$

If you let u = sin(x), du = cos(x)dx, you get an antiderivative of (1/2)u2 + C = sin2(x) + C1

OTOH, if you let u = cos(x), du = -sin(x)dx, and you get an antiderivative of -(1/2)u2(x) + C2 = -(1/2)cos2(x) + C2.

Even though the two answers appear to be significantly different, they differ by a constant (1/2). The key is the identity sin2(x) + cos2(x) = 1.