Integral of $\int 1/x dx$

Is this a sufficient proof for this integral?:

$$\int \frac{dx}{x} = \ln |x| + \mathcal{C}$$

Let $$x = e^{u} : $$

$$\int \frac{dx}{x} = \int du = u + \mathcal{C} = \ln |x| + \mathcal{C}$$

I'm not sure :S I don't know if my logic's a bit wishy washy/ circular with this :S And when I've looked, most websites seem to say it's a definition rather than a result but...I guess I just want a proof for this. Anyone have any ideas/ can validate whether the above is correct?

Thanks very much

Solution 1:

Your proof is valid but you can add more precision:

• if $x\in(0,+\infty)$ we pose $x=e^u$ and we find $$\int \frac{dx}{x} = \ln x+ \mathcal{C}$$
• if $x\in(-\infty,0)$ we pose $x=-e^u$ and we find $$\int \frac{dx}{x} = \ln (-x) + \mathcal{C}$$ then we conclude.

Solution 2:

Looks fine to me. By Taylor's Theorem, you can also write

$$\frac{1}{x} = \sum_k (-1)^k (x-1)^k$$

and integrate in the radius of convergence term by term, recognizing the result on the right-hand side. Assuming $x > 0$,

$$\begin{split} \int \frac{dx}{x} &= \int \sum_{k=0}^\infty (-1)^k (x-1)^k dx \\ &= \sum_{k=0}^\infty (-1)^k \int (x-1)^k dx \\ &= \sum_{k=0}^\infty \frac{(-1)^k}{k+1} (x-1)^{k+1} + C \\ &= \ln x +C \end{split} $$