When we assume that $$1+\log x = t,$$ then the integral becomes $$\log(1+\log x)+ C.$$
But when we assume that $$\log x = t,$$ the integral becomes
$$∫\frac1x + \frac{\log x}x\,\mathrm dx \\
\log x + \frac{(\log x)^2}2 + C.$$
Why the discrepancy?
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3$\begingroup$ Wouldn't your first integral then be $\int t dt = t^2/2+C$? $\endgroup$– RandallFeb 20 at 18:27
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$\begingroup$ Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. $\endgroup$– Another UserFeb 20 at 18:33
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1 Answer
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If you do the substitution $t=1+\log{x}$, the integral becomes$$\int t dt=\frac{1}{2}t^2+C=\frac{1}{2}(1+\log{x})^2+C=\frac{1}{2}(\log{x})^2+\log{x}+\frac{1}{2}+C$$ Combining $\frac{1}{2}$ and C into a new arbitrary constant gives you the same result
