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Today,when I found some problem on wiki,and find the integral,and I suddenly saw one of these nice problem.

$$\int\dfrac{x^2+2x+1+(3x+1)\sqrt{x+\ln{x}}}{x\sqrt{x+\ln{x}}(x+\sqrt{x+\ln{x}})}dx$$

This answer is

$$2(\sqrt{x+\ln{x}}+\ln{(x+\sqrt{x+\ln{x}})}+C$$

I try it,But I can't,and this problem is from famous wiki: http://en.wikipedia.org/wiki/Risch_algorithm#cite_note-2

so I think this is famous problem,so maybe have some nice methods,But I don't have.Thank you for your help.

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If you already know the antiderivative then you can easily get a step by step integration by doing the following:

  1. Take derivative of the antiderivative. (Step by step)
  2. Every time you use chain rule that will give you a change of variable. Each derivation of a product a partial integration, ...

In our case $$\begin{align}D\left(2(\sqrt{x+\ln(x)}+\ln(x+\sqrt{x+\ln(x)})\right)&=2D\left(\sqrt{x+\ln(x)}\right)+2D\left(\ln(x+\sqrt{x+\ln(x)})\right)\\ &=\frac{D\left(x+\ln(x)\right)}{\sqrt{x+\ln(x)}}+2\frac{D\left(x+\sqrt{x+\ln(x)}\right)}{x+\sqrt{x+\ln(x)}}\\ &=\frac{1+1/x}{\sqrt{x+\ln(x)}}+2\frac{1+D\left(\sqrt{x+\ln(x)}\right)}{x+\sqrt{x+\ln(x)}}\\ &=\frac{1+1/x}{\sqrt{x+\ln(x)}}+2\frac{1+\frac{D\left(x+\ln(x)\right)}{\sqrt{x+\ln(x)}}}{x+\sqrt{x+\ln(x)}}\\ &=\frac{1+1/x}{\sqrt{x+\ln(x)}}+2\frac{1+\frac{1+1/x}{\sqrt{x+\ln(x)}}}{x+\sqrt{x+\ln(x)}}\end{align}$$

This last simplifies/(should simplify if I didn't make some mistake differentiating) to the integrand using algebraic manipulations.

So, we can work it out backwards.

  1. We do the manipulations to transform the integrand into the right hand side in the chain of equalities above.
  2. Integration now goes the same path as the derivation $D$, but backwards, and using the replacements of the properties of derivation by the corresponding of integration mentioned above.
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    $\begingroup$ There is a $\LaTeX$ issue that must be fixed. $\endgroup$ – user93957 Oct 23 '13 at 13:05
  • $\begingroup$ Life is too short to bother. Mathematics is made for humans (at the moment) and they are good at understanding the big picture. $\endgroup$ – OR. Oct 24 '13 at 13:17

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