How can I calculate $$ \int{\sec\left(x\right)\tan\left(x\right)  \over 3x + 5}\,{\rm d}x $$  My Try:: $\displaystyle \int \frac{1}{3x+5}\left(\sec x\tan x \right)\,\mathrm dx$  Now Using Integration by Parts::  We Get $\displaystyle  = \frac{1}{3x+5}\sec x +\int  \frac{3}{(3x+5)^2}\sec x\,\mathrm  dx$  Now My Question is How Can I calculate (II) Integral.  Please explain this to me.
Cruising the old questions I came across juantheron asking for $\int \frac {\sec x\tan x}{3x+5}\,dx$ He tried using $(3x+5)^{-1}$ for $U$ and $\sec x \tan x$ for $dv$while integrating by parts.  below is his work.  How can I calculate $$ \int {\sec\left(x\right)\tan\left(x\right)  \over 3x + 5}\,{\rm d}x $$  My Try:: $\displaystyle \int \frac{1}{3x+5}\left(\sec x\tan x \right)\,\mathrm dx$  Now Using Integration by Parts::  We get  $$= \frac{1}{3x+5}\sec x +\int  \frac{3}{(3x+5)^2}\sec x\,\mathrm  dx$$  Here he hit his road block.   I tried the opposite tactic   Taking the other approach by parts.   let $$U= \sec x \tan x$$ then$$ du= \tan^2 x \sec x +\sec^3 x$$ and $$dv=(3x+5)^{-1}$$ then $$v=\frac 1 3 \ln(3x+5)$$ Thus $$\int \frac {\sec x \tan x}{3x+5}\,dx= \frac {\ln(3x+5)\sec x \tan x}{3} - \int \frac {\ln(3x+5) [\tan^2 x \sec x +\sec^3 x]}{3} \,dx$$    As you can see I got no further than he did.   So how many times do you have to complete integration by parts to get the integral of the original $\frac {\sec x \tan x}{3x+5} \, dx$ or is there a better way?
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