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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.

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    $\begingroup$ One can simply observe that the integrand is just the derivative of $\frac{x(x+2) \ln x - (x+1) \left ( (x+1) \ln (x+1) - 1 \right )}{2(x+1)^2}+C$ and obtain the result from there. $\endgroup$ – Trogdor Sep 30 '15 at 7:42
  • $\begingroup$ You're very close but not perfecrty correct. Cause this is not what I got through by parts . But anyways thanks you gave me another approach. $\endgroup$ – Adesh Tamrakar Sep 30 '15 at 7:45
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    $\begingroup$ alternatively to the nice answers below: (1) Change $x\rightarrow t-1$, (2) expand log in a taylorseries, (3) integrate termwise (4) use partial fractions and resum. $\endgroup$ – tired Sep 30 '15 at 10:01
  • $\begingroup$ can we just get the solution through sustitution method. $\endgroup$ – Adesh Tamrakar Sep 30 '15 at 13:39
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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, $$ \int \frac{\log x}{(1+x)^3}\,dx=\frac{1}{2}\Bigl(\log x+\frac{1}{1+x}-\log(1+x)-\frac{1}{(1+x)^2}\log x\Bigr)+C. $$

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  • $\begingroup$ Great answer. I have a dought that why we cannot just solve it through substitution method. $\endgroup$ – Adesh Tamrakar Sep 30 '15 at 13:40
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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 $$ \int \dfrac{\ln x}{(1+x)^3} = \dfrac{1}{2(1+x)} - \dfrac{\ln(1+x)}{2} + \left(\dfrac{1}{2} - \dfrac{1}{2(1+x)^2}\right) \ln(x) + C $$

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  • $\begingroup$ Interesting approach, but how do you come up with that conjectured form? What about the integrand suggests a "linear" combination of $\ln(1+x)$, $\ln x$, and a constant would work? $\endgroup$ – user170231 Oct 2 '15 at 14:03
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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.

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