Evaluating integral of fraction with logarithmic terms How can I formally evaluate the following integral?$$\int_{1}^5 \frac{\log(7-x)}{\log((7-x)(x+1))}dx.$$
I tried the substitution $t=\log(7-x)$ and the integrand reduces to 
$$\frac{-te^{t}}{t+\log(8-e^{t})},$$
although this does not look much simpler.  According to Wolfram, the answer appears to be $2$.  I am not sure how this can be obtained; perhaps there is some elegant trick that I am missing, because the result is unexpectedly simple.
 A: Let $$I = \int_1^5\frac{\ln(7-x)}{\ln\left[(7-x)(x+1)\right]} \, dx = \int_1^5 \frac{\ln(7-x)}{\ln(7-x)+\ln(x+1)} \, dx\tag 1$$
Now Using Formula $$\bullet \int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx$$
So $$I = \int_1^5 \frac{\ln[7-(1+5-x)]}{\ln[7-(1+5-x)]+\ln[(1+5-x)+1]} \, dx$$
So $$I= \int_1^5 \frac{\ln(x+1)}{\ln(7-x)+\ln(x+1)} \, dx \tag 2$$
Now Add $(1)$ and $(2)\;,$ We get
$$2I = \int_1^5 1 \, dx = 4\Rightarrow I = 2$$
A: $7$ is $2$ more than $5$ and $-1$ is $2$ less than $1$, and you have an integral from $1$ to $5$ involving $7-x$ and $x-(-1)$.  That suggests looking at the midpoint between $1$ and $5$, which is $3$.
\begin{align}
u & = x-3 \\[8pt]
4+u & = x + 1 \\
4-u & = 7-x
\end{align}
\begin{align}
\int_1^5 \frac{\log(7-x)}{\log(7-x) + \log(x+1)} \, dx = \int_{-2}^2 \frac{\log(4-u)}{\log(4-u)+\log(4+u)} \, du
\end{align}
and
\begin{align}
& \int_{-2}^2 \frac{\log(4-u)}{\log(4-u)+\log(4+u)} \, du + \int_{-2}^2 \frac{\log(4+u)}{\log(4-u)+\log(4+u)} \, du \tag 1 \\[10pt]
= {} & \int_{-2}^2 1\, du = 4.
\end{align}
If we can show the two integrals in $(1)$ are equal, then that and the fact that they add up to $4$ means they are each equal to $2$.
Now let
\begin{align}
w & = -u, \\
dw & = -du.
\end{align}
Then
$$
\int_{-2}^2 \frac{\log(4-u)}{\log(4-u)+\log(4+u)} \, du = \int_2^{-2} \frac{\log(4+w)}{\log(4+w)+\log(4-w)} \, (-dw) 
$$
and we're done.
