Question releating to the $\int^x_1\frac{\ln(t)}{t+1}$ 
If $f(x)=\int^x_1\frac{\ln(t)}{t+1}dt$ if $x > 0$. Compute $f(x) + f(1/x)$. As a check, you should obtain $f(2)+f(1/2)=(\ln2)^2$

I have tried evaluating the integral $\int^x_1\frac{\ln(t)}{t+1}dt$ by parts, but it turns out to be futile as $\int\frac{\ln(t)}{t+1}dt=\ln(t)\ln(t+1)-\int\frac{\ln(t+1)}{t}t$ and $\int\frac{\ln(t+1)}{t}=\ln(t)\ln(t+1)-\int\frac{\ln(t)}{t+1}dt$ which leads to $\int\frac{\ln(t)}{t+1}dt=\int\frac{\ln(t)}{t+1}dt$.
Is there any other way to evaluate this integral?
 A: Let $g(x)=f(x)+f(1/x)$ for $x>0$. It is straightforward to check that
$$\eqalign{
g'(x)&=f'(x)-\frac{1}{x^2}f'\left(\frac{1}{x}\right)\cr
&=\frac{\ln x}{1+x}+\frac{\ln x}{x(1+x)}=\frac{\ln x}{x}\cr
&=\frac{1}{2}\left(\ln^2x\right)'}
$$
This proves that $x\mapsto g(x)-\frac{1}{2}\ln^2x$ is constant on $(0,+\infty)$ because it has a zero derivative. This constant is zero, because this function vanishes for $x=1$. Thus $$\forall\, x>0,\qquad g(x)=\dfrac{1}{2}\ln^2x.$$ 
That is 
$$\forall\, x>0,\qquad f(x)+f\left(\frac{1}{x}\right)=\frac{1}{2}\ln^2 x.$$
A: What you want is
$$f(x) + f(1/x)$$
thus
$$ \int_1^x \frac{\ln(t)}{1+t} dt + \int_1^{1/x} \frac{\ln(t)}{t+1} dt $$
which can be written as
$$ \int_1^x \frac{\ln(t)}{1+t} dt + \int_1^x \frac{\ln(1/t)}{1/t+1} d[1/t] $$
Thus
$$ f(x) + f(1/x) = \int_1^x \frac{\ln(t)}{t} dt $$
And this can be calculated
$$ \int \frac{\ln(t)}{t} dt = \frac{\ln^2(t)}{2} $$
So you find
$$ f(x) + f(1/x) = \frac{\ln^2(x)}{2} $$
A: You do not have to evaluate the integral. Note that 
\begin{align*}
   f(1/x) &= \int_1^{1/x} \frac{\log t}{t+1} \, dt\\
          &= \int_1^x \frac{\log 1/\tau}{\frac 1{\tau} + 1}\,\left(-\frac 1{\tau^2}\right) d\tau & \text{substition $t = \frac 1\tau$}\\
          &= \int_1^x \frac{\log\tau}{\tau+\tau^2}\, d\tau\\
          &= \int_1^x \frac{\log \tau}{\tau}\, d\tau - \int_1^x\frac{\log\tau}{1+\tau}\, d\tau\\
          &= \int_0^{\log x} u\,du - f(x)\\
          &= \frac 12\log^2 x - f(x).
\end{align*}
