Integration by Parts, 1/x dillema

[erjkism]

the answer to $$\int\frac{1}{x}(dx)$$ is lnx +c. but if you do it with integration by parts, you end up with

$$\int\frac{1}{x}(dx)$$ = 1 + $$\int\frac{1}{x}(dx)$$

which comes to 0=1. why does this happen?

 

[Vid]

The derivation of integration by parts comes from the product rule so the (uv) term is actually $$\int(uv)'$$. The constant from this integration is usually assumed to be combined with the constant from the other integral, but since the other integrals canceled out we still have a constant term 1 + c = 0.

Though, this is just a guess, and I could be completely wrong.

 

[jostpuur]

I don't think it's possible to tell what you did wrong unless you give more details of your calculation. Did you start with something like this

$$\frac{1}{x} = \frac{1}{x} (Dx) = D\Big(\frac{1}{x}x\Big) - \Big(D\frac{1}{x}\Big)x = D\Big(\frac{1}{x}x\Big)+\frac{1}{x}?$$

But I'm not ending up into any paradox if I perform the integration by parts with this. You just get

$$\int_a^b \frac{1}{x}dx = \frac{b}{b} - \frac{a}{a} + \int_a^b\frac{1}{x}dx.$$

 

[erjkism]

im not using a definite integral in my calculation. there aren't any limits "a" and "b"

 

[Ben Niehoff]

Ah, but it's a trick! There is really no such thing as an indefinite integral. The notation

$$\int f(x) \; dx$$

is really just shorthand for

$$\int_a^x f(t) \; dt$$

for some arbitrary constant a. If you write it out like this, so that a is consistent, then you will resolve the paradox.

Alternatively, you could be absolutely pedantic about your arbitrary constant C:

$$\int \left( \frac{d}{dx} f(x) \right) dx = f(x) + C$$

and you'll note that it is certainly true that

0 = 1 + C

for some C.

 

[jostpuur]

im not using a definite integral in my calculation. there aren't any limits "a" and "b"

In a sense, the integral functions are definite integrals too. If F(x) is chosen so that its derivative is f(x), then it is

$$F(x) = \int_{x_0}^x f(u) du$$

with some $x_0$.

Your problem seems to be, that you are using a formula

$$\int fg' dx = fg - \int f'g dx$$

without knowing what it means. My advice is that try to understand how the formula is derived, and everything should get clearer.

It seems somebody was faster than me.