Compute $\int_{0}^{\pi/4}\ln(1-\sqrt[n]{\tan x})\frac{dx}{\cos^2(x)}$ I am trying to compute this
$$
\int_{0}^{\pi/4}\ln(1-\sqrt[n]{\tan x})\frac{\mathrm dx}{\cos^2(x)},\qquad
(n\ge1).
$$
Making a transformation of $I$ to utilise a sub of $u=1-\sqrt[n]{\tan x}$
\begin{align}
I&=\int_{0}^{\pi/4}\frac{\sec^2(x)}{n\sqrt[n]{\tan x}}\cdot n\sqrt[n]{\tan x}\cdot\ln(1-\sqrt[n]{\tan x})\,\mathrm dx
\\[6px]
&\qquad\mathrm dx=-\frac{n\sqrt[n]{\tan x}}{\sec^2(x)}\,\mathrm du
\\[6px]
I&=n\int_{0}^{1}(1-u)^{n-1}\ln u \,\mathrm du
\end{align}
This can be easily done by integration by parts, but I seem to shruggle in somewhere in evaluating it,
$$
\int(1-u)^{n-1}\ln u \,\mathrm du=
n(1-u)^n\ln u-\frac{1}{n^2}\int \frac{(1-u)^n}{u}\,\mathrm du
$$

Your integral is given by the negative of the  $n$-th harmonic number:
$$ I_n \equiv n \int \limits_0^1 (1-u)^{n-1} \ln (u) \, \mathrm{d} u = - H_n = - \sum \limits_{k=1}^n \frac{1}{k} \, . $$
You can use the substitution $u = 1-t$ and then have a look at this question for a derivation. Here's an alternative route: 
Use the antiderivative $\frac{t^n - 1}{n}$ of $t^{n-1}$ to integrate by parts directly:
\begin{align} 
I_n &= n \int \limits_0^1 t^{n-1} \ln (1-t) \, \mathrm{d} t = - \int \limits_0^1 \frac{1-t^n}{1-t} \, \mathrm{d} t = - \sum \limits_{k=0}^{n-1} \int \limits_0^1 t^k \, \mathrm{d} t = -\sum \limits_{k=0}^{n-1} \frac{1}{k+1} = - H_n \, .
\end{align}
