When is the integral of $x^{-1} = \ln \lvert x \rvert$?

Question

In my textbook there is a question that says:

Use either a suitable substitution or the reverse chain rule to find the following integrals.

$$ \int \left(\frac{x}{\sqrt{x^2 + 2}}\right) dx$$

Using substitution I came down to:

$${1\over 2} \int u^{-{1\over2}}du$$ where $u=x^2+2$

Couldn't this be rewritten as $${1\over 2} \int (\sqrt{u})^{-1}$$ And isn't that equivalent to $${1\over 2}\log|\sqrt{u}| + c$$ where $\sqrt{u}$ acts as $x$?

The thing is the answer is $\sqrt{x^2 +2} + c$. And so I don't know if I fully understand the theory behind integrating $1/x$ and $\log|x|$.

Thanks!

Answer 1

$\int \frac {1}{u} du = \ln |u|+C.$ However, $\int \frac {1}{f(u)}\ du \ne \ln |f(u)|$

That is, you can only integate into a logarthm if, by u-substitutions you can get the denominator with that lone $u.$

For this problem we have a power rule you can employ:

$\int u^a du = \frac {u^{a+1}}{a+1} + C$ (which works for all constants $a\ne -1$ )

Answer 2

We have that

• for $x>0$ $$f(x)=\ln |x|=\ln x\implies f’(x)=\frac1x$$

• for $x<0$ $$g(x)=\ln |x|=\ln (-x)\implies g’(x)=-\frac1{-x}=\frac1x$$