$\int\frac{x}{(x\tan x+1)^2}\,\mathrm{d}x$

Asked Feb 10 '18 at 02:00

Active Feb 10 '18 at 02:27

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$$ \int\frac{x}{(x\tan x+1)^2}\,\mathrm{d}x = \int x \frac{\cot^2 x}{(x+\cot x)^2}\,\mathrm{d}x.$$

By parts method gives $$-\frac{x}{x+\cot x}+\int\frac{1}{x+\cot x}\,\mathrm{d}x,$$ and how to solve it?

edited Feb 10 '18 at 02:27

Ѕᴀᴀᴅ

asked Feb 10 '18 at 02:00

jacky

1

While I am on a time crunch, something tells me that you need to "reverse" the quotient rule to get back to the original function. Why? Because the denominator is squared. – imranfat Feb 10 '18 at 02:28

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