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I came across a question today:

$$\int \frac{\ln x \, dx}{(1+\ln x)^2}$$

How to do it? I tried to substitute $1+\ln x$ and also tried taking $\ln x$ outside the brackets but they doesn't work... I even tried to use that multiplication rule.That didn't work either. I can't see anything to pass through it...

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  • $\begingroup$ @Travis Edited.. $\endgroup$ – manshu Mar 1 '16 at 14:12
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Let $u=\ln x$ and $du=\frac{1}{x}dx$.

Thus, the integral transforms to $\displaystyle\int \dfrac{\ln x dx}{(1+\ln x)^2}=\displaystyle \int \frac{e^u u du}{(1+u)^2}.$

Now, integrate by parts with $f=e^u u$ and $dg=\frac{1}{(1+u)^2}du.$

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Hint: Wolfram Alpha solves this (in an admittingly cumbersome way) by first letting $v=\log x$ and then integrating by parts with $u = ve^v$ and $dg = \frac{dv}{(v+1)^2}$. Then just back substitute and you'll get your answer.

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Set $t=\ln x$ and $dt=\frac{dx}{x}$

$$=\int\frac{e^tt}{(t+1)^2}dt$$

By parts $f=e^tt$ and $dg=\frac{dt}{(1+t)^2}$

$$=-\frac{e^tt}{t+1}+\int e^t dt=\color{red}{\frac{x}{\ln x+1}+\mathcal C}$$

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  • 1
    $\begingroup$ I love this integral. Integration by parts works so perfectly here. $\endgroup$ – zz20s Mar 1 '16 at 15:09
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$1+\ln x=u\iff x=e^{u-1},dx=e^{u-1}du$

$$I=\int\dfrac{\ln x}{(1+\ln x)^2}dx=\int\dfrac{u-1}{u^2}e^{u-1}du=\int e^{u-1}\left(\dfrac1u+\dfrac{d(1/u)}{du}\right)du =\dfrac{e^{u-1}}u+K$$

as $\dfrac{e^xf(x)}{dx}=e^x[f(x)+f'(x)]\implies\int e^x[f(x)+f'(x)]dx=e^x f(x)$

$$\implies I=\dfrac x{1+\ln x}+K$$

Alternatively,

$$\dfrac{\ln x}{(1+\ln x)^2}=\dfrac{1+\ln x-1}{(1+\ln x)^2}$$

$$=\dfrac1{1+\ln x}\cdot\dfrac{dx}{dx}+x\cdot\dfrac{d\{1/(1+\ln x)\}}{dx}$$

What is $\displaystyle\dfrac{d(uv)}{dx}=?\implies\int(u\dfrac{dv}{dx}+v\dfrac{du}{dx})dx=uv+K$

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