What is the mistake in doing integration by this method?

Question

Integration Of a given function can be found out in many ways, For a specific function ∫1/xlogx, if we do integration by parts (∫f(x) g(x)= f(x) ∫ g(x)- ∫ [d/dx (f(x)) ∫g(x)] dx ) we get this way

Let ∫1/xlogxdx = A

If we do integration by parts, we have

$$\int \frac{1}{x\log x} \, dx= \frac{1}{\log x}\cdot\int \frac{1}{x} \, dx - \int \frac{d}{dx}\left(\frac{1}{\log x}\right)\,dx\cdot\int \frac{1}{x} \, dx+c$$ $$=\frac{1}{\log x}\cdot \log x+\int \frac{1}{x\log x}\,dx+c$$

i.e, A = 1 + A + c

Which tells us that the value of "$c$" in this specific integral is $-1$. So does this mean that the Integral of function $\dfrac{1}{x\log x}$ has only one value.! So this means only two things, either I am wrong somewhere or i am missing a point somewhere.

Edit: I do know we can simple get it by writing logx as u and ¹/x as du, we directly get it as log(logx) +c.

But I wanted to know the reason why we can't apply By parts when we can integrate it by substitution.

Answer 1

The indefinite integral is only unique up to a constant. Thats why you have $$\int{\frac{1}{x\log(x)}dx} = 1+\int{\frac{1}{x \log(x)}dx}$$ If you differentiate on both sides you get $$\frac{1}{x\log(x)} = \frac{1}{x\log(x)}$$ Your $c$ doesn't has to be $-1$, it can be any value because generally $$\int{\frac{1}{x\log(x)}dx} - \int{\frac{1}{x \log(x)}dx} \neq 0$$

Answer 2

You shouldn't be doing integration by parts here since $\frac1x$ is the derivative of $\ln x$. Instead do the substitution $u=\ln x$

Without wishing to generalize too much, you need to distinguish between integrands of the form $$f(x)g(x)$$ where $f$ and $g$ are completely different types of function, i.e. not related by differentiation, and integrands of the form $$g'(x)f'(g(x))$$ where the substitution $u=g(x)$ is made (as is the case in your example)

Answer 3

A substitution of $u = \log x$ here yields $x \, \mathrm{d}u = \mathrm{d}x$. So you get $$\int \frac{1}{u} \, \mathrm{d}u = \log u + \mathrm{C}$$

Back-substitution yields $$\bbox[border: 1px solid blue, 10px]{\int \frac{\mathrm{d}x}{x \log x} = \log \log x + \mathrm{C}.} $$

Answer 4

The real answer is that you did integration by parts and basically got $x=x$. That doesn't mean there is just a constant value for your integral. It just meant that your integration by parts looped around and did nothing - the constant $+1$ can just be ignored because you are dealing with indefinite integrals.

So, you can apply integration by parts, you just don't get any closer to the solution. It's a null operation. You can sometimes get caught in the same trap if you try to do integration by parts twice and accidentally undo the previous step. Sometimes a simplification just doesn't simplify anything and you have to find another way. Just don't confuse this with the other (desired) effect when you get the same integral back, but in a way that you can actually solve the equation from it (in those cases you don't get the same thing back, but express it with the same integral, but in a different way, not as the meaningless tautology).

Answer 5

There is a problem with the integration-by-parts formula, often written

$$ \ \int \ \ u \ dv \ = \ u \ v \ - \ \int \ \ v \ du \ \ , $$

for certain functions. If it should happen that $ \ v \ = \ \frac{c}{u} \ $ and $ \ u \ dv \ = \ - v \ du \ $ , then the method is not going to take us anywhere. The second condition gives us a differential equation

$$ \ \frac{dv}{v} \ = \ -\frac{du}{u} \ \ \Rightarrow \ \ \log v \ = \ -\log u \ = \ \log(\frac{1}{u}) \ \ , $$

which is unfortunately satsified by $ \ u \ = \ \frac{1}{\log x} \ \ , \ \ v \ = \ \log x \ \ \Rightarrow \ \ dv \ = \ \frac{1}{x} \ dx \ \ $ .