Variables of integration. (originally a picture from a textbook:)

$\displaystyle f(x) = \int_1^x \frac {\ln t}{1 + t} \, \mathrm dt \text{ for } x > 0$

Now,

$\displaystyle f(1/x) = \int_1^{1/x} \frac {\ln t}{1 + t} \, \mathrm dt$

Put

$t = \dfrac 1 u \ \implies \ \mathrm dt = - \dfrac 1 {u^2} \, \mathrm du$
$\therefore \ f(1/x) =  \displaystyle \int_1^x \frac {\ln(1/u)}{1 + 1/u}\frac{-1}{u^2} \, \mathrm du$
$= \displaystyle \int_1^x \frac {\ln u}{u(u+1)}\, \mathrm du = \displaystyle \int_1^x \frac {\ln t}{t(t+1)}\, \mathrm dt$

Now,

$\displaystyle f(x) + f\left({\frac 1 x }\right) =  \int_1^x \frac {\ln t}{1 + t} \, \mathrm dt + \int_1^x \frac {\ln t}{t(t+1)}\, \mathrm dt$
$\displaystyle = \int_1^x \frac {(1 + t)\ln t}{t(1 + t)} \, \mathrm dt$
$\displaystyle = \int_1^x \frac {\ln t}{t} \, \mathrm dt$
$\displaystyle = \left.{\frac 1 2 \left({\ln t}\right)^2}\right\vert_1^x = \frac 1 2 \left({\ln x }\right)^2$

My question is:
Here $t$ is taken to be $1/u$. But later $u$ is replaced by $t$ itself. I understand that integration is independent of the variable chosen but won't this imply that $t=1/t$? Where am I getting wrong?
 A: Perhaps it would have been better to keep it as a $u$ so as to avoid the confusion you're having, or to use a third variable.
When you have a function of the form:
$$\phi(x) = \int_{c}^x f(\nu) \, \mathrm d\nu$$
then  $\nu$ has no importance other than to tell you that you the function is an antiderivative of the integrand (whether you can solve the integral or not), and the variable used inside the integrand is of no importance. The function takes as an argument $x$. You can think of the line that's confusing you as a different $t$.  Or you can think of it as a sort of "temporary variable" (sometimes called a "dummy variable") that doesn't appear once $\phi(x)$ is evaluated anyway, so we don't care what symbol we use: $\nu, t, \star, \text{ etc. }$
You could have combined the integrals even if one is written in terms of $u$ and one in terms of $t$, because what the variable in the integrand "looks like" doesn't matter. When the book says "let $t = 1/u$ etc, think of it this way:

"We replace $t$ with $1/t$ in the integrand, and adjust the differential and bounds of integration according to the rules of integration by substitution"

Replacing $t$ with $1/t$ doesn't mean $t = 1/t$, it means that we're rewriting the integral in a different way. The author decided that "let $t = 1/u$ was less confusing than "replace $t$ with $1/t$".
The upshot: It does not matter what the variable of integration "looks like", because the integral is defined as a function of the bound(s) , and we just need some variable in the integrand because we don't want it to look like this:
$$f(x) = \displaystyle \int_1^x \frac {\ln \phantom{t}}{1 + \phantom{t}} \, \mathrm d\phantom{t}$$
A: Note that dummy variables can have any name:
$$\int_a^b f(\star)\>d\star=\int_a^b f(\bullet)\>d\bullet\ .$$
These are "internal variables" for the system that does the computation for you. The result will be a number which depends only on the function $f$ and on the given boundaries $a$ and $b$.
