Solving $\int\frac{1+\ln x}{x^2\ln^2 x}dx$ by method of parts 
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*I need solve the following integral:
$$\int\frac{1+\ln x}{x^2\ln^2 x}dx,$$
by the method of parts. 
My attempt:
\begin{align*}
 \int\frac{1+\ln x}{x^2\ln^2 x}dx={}&\int\frac{dx}{x^2\ln^2 x}dx+\int\frac{\ln x}{x^2\ln^2 x}dx=\int\frac{dx}{x^2\ln^2 x}dx+\int\frac{dx}{x^2\ln x}={} \\
 {}={}&\int{x^{-2}\ln^{-2}dx}+\int{x^{-2}\ln^{-1}dx}
 \end{align*}
Now we have a substution:
$$u=\ln^{-2}x\Rightarrow du=-2\ln^{-3}x\cdot\frac{1}{x}dx,$$
etc., etc.
My questions is: I do not know a good deal, or should some other way. Thanks.

*And, I need solve the integral:
$$\int\frac{\ln^2(a+bx)}{x^n}dx.$$
I can solve the integral $\int\frac{\ln(a+bx)}{x^n}dx$ by the method of parts, by I don't know how to solve integral 2 by the method of parts. Thanks.
Thanks for your help. 
 A: Substitute $\displaystyle u = \frac{1}{x \ln{x}}$ and $\displaystyle du = -\frac{\ln{x}+1}{x^2 \ln{x}^2}dx$.
This directly gives you the solution.
A: Let $$\displaystyle I = \int\frac{1+\ln x}{x^2(\ln x)^2}dx\;,$$ Let $\ln(x) = t\Rightarrow x=e^t\;,$ Then $dx = e^tdt$
So Integral $$\displaystyle I = \int \frac{1+t}{e^{2t}\cdot t^2}\cdot e^t dt = \int \left(t^{-1}+t^{-2}\right)\cdot e^{-t}dt = \int t^{-1}\cdot e^{-t}dt+\int t^{-2}\cdot e^{-t}dt$$
Now Integration by parts for First Integral
$$\displaystyle I = -t^{-1}\cdot e^{-t}-\int t^{-2}\cdot e^{-t}dt+\int t^{-2}\cdot e^{-t}dt+\mathcal{C} = -\frac{1}{t\cdot e^{t}}+\mathcal{C}$$
So we get $$\displaystyle I =\int\frac{1+\ln x}{x^2(\ln x)^2}dx= -\left[\frac{1}{x\cdot \ln x}\right]+\mathcal{C}$$
OR we can write $$\displaystyle \frac{1+\ln x}{(x\cdot \ln x)^2} = \frac{x \ln x\cdot (1)'-(1)\cdot (x\ln x)'}{(x\cdot \ln x)^2} = -\frac{d}{dx}\left[\frac{1}{x\cdot \ln x}\right]$$
Same as MJay1985.
A: HINT: Let $\ln x=t\implies \frac{dx}{x}=dt$ $$\int\frac{1+\ln x}{x^2(\ln x)^2}dx$$
$$=\int\frac{1+t}{e^t(t)^2}dt$$
$$=\int e^{-t}\left(\frac{1}{t^2}+\frac{1}{t}\right)dt$$
I hope you can take it from here
A: Integration by parts works quite well here.  
First, we proceed to split the integral as proposed in the OP and write
$$\int\frac{1+\log x}{x^2\log^2 x}\,dx=\int\frac{1}{x^2\log^2 x}\,dx+\int\frac{1}{x^2\log x}\,dx \tag 1$$
Then, we integrate by parts the first integral on the right-hand side of $(1)$.  Here, we let $u=\frac1x$ and $v=-\frac{1}{\log x}$.  Then, $(1)$ becomes
$$\int\frac{1+\log x}{x^2\log^2 x}\,dx=-\frac{1}{x\log x}-\int\frac{1}{x^2\log x}\,dx+\int\frac{1}{x^2\log x}\,dx=-\frac{1}{x\log x}+C$$
Therefore, we have 
$$\bbox[5px,border:2px solid #C0A000]{\int\frac{1+\log x}{x^2\log^2 x}\,dx=-\frac{1}{x\log x}+C}$$
