How can I calculate $\displaystyle \int \frac{\sec x\tan x}{3x+5}\,\mathrm dx$
My Try:: $\displaystyle \int \frac{1}{3x+5}\left(\sec x\tan x \right)\,\mathrm dx$
Now Using Integration by Parts::
We Get $\displaystyle = \frac{1}{3x+5}\sec x +\int \frac{1}{(3x+5)^2}\sec x\,\mathrm dx$
Now My...
The question we discussed was so ready-made . :P None would think of substituting x=tan y, unless it is Chris's Wise sister(The question was asked by him/her, ironic).:P
I know, I have been there. Didn't sign up. Looked pretty boring.:P
the forum is boring... but Kunny's integral collection is the best there
By the way, whenever you see $1+x^2$, there is a pretty natural itch to substitute $x = tan \theta$, so I would say that the substitution was ready made as well :P