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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.

Archer
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juantheron
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0 Answers0