$$\int \frac{1}{x+x\log x}\,dx$$
I couldn't use any of the integration techniques to solve this, any help will be appreciated!
Find $\int \frac{1}{x+x\log x}dx$
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$\begingroup$
calculus
integration
2 Answers
5
"HINT": (Assuming of course that $\log=\ln$) It is $$\int\frac{1}{x}\frac{1}{1+\ln x}\,dx$$ so you can make the substitution $y=\ln x$.
3
$$ \int \frac{1}{x}\frac{1}{1+\log x}dx $$ let $1+ \log x = u\implies du =\frac{1}{x}\frac{1}{\ln 10}dx$
then we have $$ \int \ln 10\frac{1}{u}du = \ln 10 \ln u + C $$
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0I should add that this answer assumes $\log$ is the common log. – 2017-02-10
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0@LanierFreeman Yes, because I was not sure if the user was talking about natural or base 10. In either case there are two answers covering both :). – 2017-02-10