int9^(log(3)(secx))dx=...

`int9^(log_(3)(secx))dx=`

A

(a) `secx.tanx+c`

B

(b) `cotx+c`

C

(c) `tanx+c`

D

(d) `sec^(2)x+tan^(2)x+c`

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The correct Answer is:

C

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int(log(3x))/(xlog(9x))*dx=

A

`log(3x)-log(9x)+c`

B

`log(x)-(log3)*log(log9x)+c`

C

`log9-(logx)*log(log3x)+c`

D

`log(x)+(log3)*log(log9x)+c`

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