Integral $\int_0^1\log(1+x)\frac{1+x^2}{(1+x)^4}dx=-\frac{\log 2}{3}+\frac{23}{72}$ EDIT:  Small Typo Fixed now, Thanks to Sir Chen Wang!
Hi I am trying to prove this result without using a series approach 
$$
\int_0^1\log(1+x)\frac{1+x^2}{(1+x)^4}dx=-\frac{\log 2}{3}+\frac{23}{72}.
$$I know we can just solve it by writing
$$
\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}\left( \int_0^1\frac{x^n}{(1+x)^4}dx+\int_0^1 \frac{x^{2+n}}{(1+x)^4}dx\right),
$$
which leads to summing a bunch of Harmonic numbers which is not so easy.
This method is brute force but I often have trouble summing harmonic numbers, can we prove it another way?
Thanks
 A: Hint: Write $\ln(a+x)$ and then differentiate with regards to a. Your integrand becomes a rational function. Observation: Please note that by multiplying the end result by $3$, we have $\dfrac{69}{72}-\ln2$. Since the integrand is obviously positive, this leads me to believe that this integral was originally meant as an exercise, much similar to this one, in showing that $\ln2<\dfrac{23}{24}$ . The choice of numbers is especially interesting since, on one hand, $\ln2\approx69\%$, and on the other hand, the famous rule of $72$, still used in economics until today, is also connected to the same transcendental quantity, $\ln2$.
A: $$
\begin{align*}
I&=\int^1_0\log(1+x)\frac{1+x^2}{(1+x)^4}dx\\
&=-\frac13\int^1_0\log(1+x)d\left(\frac{2+3x+3x^2}{(1+x)^3}\right)\\
&=-\frac13\left(\left.\log(1+x)\frac{2+3x+3x^2}{(1+x)^3}\right|^1_0-\int^1_0\frac{2+3x+3x^2}{(1+x)^3}d\,\log(1+x)\right)\\
&=-\frac{\log 2}{3}+\frac13\int^1_0\frac{2+3x+3x^2}{(1+x)^4}dx\\
&=-\frac{\log 2}{3}+\frac13\int^1_0\left(\frac{2}{(1+x)^4}-\frac{3}{(1+x)^3}+\frac{3}{(1+x)^2}\right)dx\\
&=-\frac{\log 2}{3}+\frac13\left(\frac{7}{12}-\frac{9}{8}+\frac{3}{2}\right)\\
&=-\frac{\log 2}{3}+\frac{23}{72}.
\end{align*}
$$
