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Since beating my head against a brick wall is so fun, I kept working on this old integral $\int \frac{ \sec{x} \tan{x}}{3x+5}dx$ . I think I have finally found a way to do it. Here goes. $$ \int \frac{ \sec{x} \tan{x}}{3x+5} $$ $$\text {take the nat. log then} $$ $$ \int ln \frac{ \sec{x} \tan{x}}{3x+5}= \int ln \sec x+ \int ln \tan x - \int ln(3x+5) dx \rightarrow $$Wolfram shows the ugliest complex expression I have ever seen (yeah I cheated so sue me). $$ \int (\log(\sec(x))+\log(\tan(x))-log(3 x+5)) dx = 1/6 (-3 i Li_2(-e^{2 i x})-3 i (Li_2(-i \tan(x))+\log(1+i \tan(x)) \log(\tan(x)))+3 i (Li_2(i \tan(x))+\log(1-i \tan(x)) \log(\tan(x)))-3 i x^2+6 x+6 x \log(1+e^{2 i x})-2 (3 x+5) \log(3 x+5)+6 x \log(\sec(x))+10)+constant $$

Problem is I still have no idea how to reconcile the natural log after integrating the expression in order to get the integral of the original expression. Is there are way to do this? Or have I just found another dead end?

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You are integration a totally different function. In some cases, we can use the log to find derivatives of some functions, since it makes it easier to find the derivative. – Mhenni Benghorbal Feb 7 '14 at 20:47
All I have done is put the original expression in the logarithmic plane. I need to know how to map the integral back to the x y plane. Correct?? – Chris Feb 7 '14 at 20:51
up vote 3 down vote accepted

Integration and the $log$ operation are not interchangeable, and moreover, there is no general way to "undo" this:

$$\int \log f(x)dx\neq g\left(\int f(x)dx\right)$$

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So then the answer is nice try....keep trying. Yeah well. – Chris Feb 7 '14 at 21:06
@Chris - that is not the answer. Do not keep trying, since as was explained in the answer to your previous question, "...this particular integral is easy to show non-integrable in elementary terms". – nbubis Feb 7 '14 at 21:11
Yeah I'll let this one slide. Thanks for the new subject to research. The Risch algorithm – Chris Feb 7 '14 at 21:14

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