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.
 A: First, I just want to note that I really don't see why one wouldn't use integration by parts. With that said:
One way is going via FTC and a double integral.
We note that
$$
\int_1^x\frac{\log t}{(1+t)^3}\,dt
$$
gives one primitive. Writing
$$
\log t=\int_1^t\frac{1}{s}\,ds
$$
we find that (changing order of integration, and evaluating the integrals)
$$
\begin{aligned}
\int_1^x\frac{\log t}{(1+t)^3}\,dt&=\int_1^x\int_1^t\frac{1}{s(1+t)^3}\,ds\,dt\\
&=\int_1^x\int_s^x \frac{1}{(1+t)^3}\,dt\frac{1}{s}\,ds\\
&=\int_1^x\Bigl[\frac{1}{2(1+s)^2}-\frac{1}{2(1+x)^2}\Bigr]\frac{1}{s}\,ds\\
&=\frac{1}{2}\int_1^x \frac{1}{s}-\frac{1}{(1+s)^2}-\frac{1}{1+s}-\frac{1}{(1+x)^2s}\,ds\\
&=\frac{1}{2}\Bigl[\log s+\frac{1}{1+s}-\log(1+s)-\frac{1}{(1+x)^2}\log s\Bigr]_1^x\\
&=\frac{1}{2}\Bigl(\log x+\frac{1}{1+x}-\log(1+x)-\frac{1}{(1+x)^2}\log x-\frac{1}{2}+\log 2\Bigr)
\end{aligned}
$$
Thus,
$$
\int \frac{\log x}{(1+x)^3}\,dx=\frac{1}{2}\Bigl(\log x+\frac{1}{1+x}-\log(1+x)-\frac{1}{(1+x)^2}\log x\Bigr)+C.
$$
A: 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
$$ \int \dfrac{\ln x}{(1+x)^3} = \dfrac{1}{2(1+x)} - \dfrac{\ln(1+x)}{2} + \left(\dfrac{1}{2} - \dfrac{1}{2(1+x)^2}\right) \ln(x) + C $$
A: I think I got more simple answer.
we can write the above as-
$$\int-(\frac{-2logx}{(1+x)^3}+\frac{1}{x(1+x)^2})+\frac{1}{x(1+x)^2}dx$$
now we know that
differentiation of 
$$\frac{logx}{(1+x)^2}$$
is $$\frac{-2logx}{(1+x)^3}+\frac{1}{x(1+x)^2}$$
therefore now we can integrate
by replacing $$logx$$ term and easily integrating the next algebraic term.
A: Integrate by parts
\begin{align}
\int \frac {\log x}{(1+x)^3}\,dx 
& = \int \frac {\log x}2d\left(1-\frac{1}{(1+x)^2}\right)\\
&=\frac12{\log x}\left(1-\frac{1}{(1+x)^2}\right)-\frac12\int\left(\frac1{1+x}+\frac1{(1+x)^2}\right)dx\\
&= \frac12\left( \frac{x(x+2)\log x}{(1+x)^2}-\log(1+x)+\frac1{1+x}\right)+C
\end{align}
