$$\int\frac{\ln(x)}{1+\ln(x)^2}\mathrm{d}x$$ I surely know the integral would be of $u/v$ type but I am not getting any good substitution to go for it. I think $\log(x)=e^t$ would be good as we get again an $e^{e^t}$ so everything now is in $e$ but I can't go further with it. Thanks!
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Your integral can not be expressed in elementary functions. It requires the use of the exponential integral. Check with WolframAlpha. – Ian Miller May 11 '16 at 14:04
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It can be done if by log(x) you mean ln(x) – bulbasaur May 11 '16 at 14:05
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@bulbasaur Most people usually use $\log x$ as $\ln x$, at least on this site. – S.C.B. May 11 '16 at 14:11
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Ya I'll edit it sorry for confusion – Archis Welankar May 11 '16 at 14:34
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2By setting $x=e^{-t}$ you may check that an exponential integral is involved, namely $$\int \left(\frac{1}{t+i}+\frac{1}{t-i}\right) e^{-t},dt,$$ that is not an elementary integral. – Jack D'Aurizio May 11 '16 at 14:45
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Okay so can you please tell me how to go for the answer – Archis Welankar May 11 '16 at 15:02
1 Answers
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Substitute $u=\ln x$ and decompose $$\int\frac{\ln x}{\ln ^2x+1}\,\mathrm d x=\int\frac{ue^u}{u^2+1}\,\mathrm d u=\frac 12\int\frac{e^u}{u+i}\,\mathrm d u+\frac 12\int\frac{e^u}{u-i}\,\mathrm du$$
These two integrals are exponential integrals, as it can be seen if you substitute for the denominators: $$\frac 12\int\frac{e^u}{u+i}\,\mathrm d u+\frac 12\int\frac{e^u}{u-i}\,\mathrm du=\frac{e^{-i}}{2}\int\frac{e^v}{v}\,\mathrm d v+\frac{e^i}{2}\int\frac{e^t}{t}\,\mathrm d t=\frac{e^{-i}\operatorname{Ei}(v)}{2}+\frac{e^i\operatorname{Ei}(t)}{2}$$
Hence, the result is: $$\int\frac{\ln x}{\ln ^2x+1}\,\mathrm d x=\frac{e^{-i}\operatorname{Ei}(\ln x+i)}{2}+\frac{e^i\operatorname{Ei}(\ln x-i)}{2}+C$$
John Doe
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