Integration of elementary function $\int \frac {\log x}{(1+x)^3}\,{\rm d}x $ 
The question is to find the integral
  $$\int \frac {\log x}{(1+x)^3}\,{\rm d}x $$

It can be easily be solved by integration by parts, but I want to solve it without using integration by parts.

It can be done by a sort of "undetermined coefficients".  Suppose we can guess that the solution is of the form
$$ F(x) = a(x) + b(x) \ln(1+x) + c(x) \ln(x) $$
where $a, b, c$ are rational functions.   Taking the derivative and comparing to $\ln(x)/(1+x)^3$, we see
$$ \eqalign{c' &= \dfrac{1}{(1+x)^3}\cr
            b' &= 0\cr
            a' &= -\dfrac{b}{1+x} - \dfrac{c}{x}\cr} $$
From the first equation, 
$$c = c_0 - \dfrac{1}{2(1+x)^2}$$
with $c_0$ constant.  From the second, $b$ is constant.
And then, using partial fractions 
$$ a' = -\dfrac{b}{1+x} - \dfrac{c}{x} = \dfrac{1-2c_0}{2x} - \dfrac{1}{2(1+x)^2} - \dfrac{1+2b}{2(1+x)}$$
In order for $a$ to be a rational function, the terms in $x^{-1}$ and 
$(1+x)^{-1}$ must vanish, so $b=-1/2$ and $c_0 = 1/2$.  Then we get
$$ \eqalign{a' &= - \dfrac{1}{2(1+x)^2}\cr a &= \dfrac{1}{2(1+x)} + C\cr}$$
so that
$$ \int \dfrac{\ln x}{(1+x)^3} = \dfrac{1}{2(1+x)} - \dfrac{\ln(1+x)}{2} + \left(\dfrac{1}{2} - \dfrac{1}{2(1+x)^2}\right) \ln(x) + C $$
