Solving $\int\frac{1+\ln x}{x^2\ln^2 x}dx$ by method of parts

Question

1. 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.

2. 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.

Answer 1

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.

Answer 2

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.

Answer 3

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

Answer 4

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

$$\bbox[5px,border:2px solid #C0A000]{\int\frac{1+\log x}{x^2\log^2 x}\,dx=-\frac{1}{x\log x}+C}$$