please help me, i got final exam tomorrow integrate : [(2+arctan 3x)^(1/2) / 1+9x^2

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please help me, i got final exam tomorrow integrate : [(2+arctan 3x)^(1/2) / 1+9x^2

Mathematics
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\[\int\limits_{}^{}(2+\tan ^{-1}3x)^{1/2}\div(1+9x ^{2})\]
if u=2+tan^-1(3x) du = 3x/(1+9x^2) dx
make that 3/(1+9x^2)

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[S] u^(1/2)/3 du
2(2+tan^-1(3x)) --------------- +C ?? 9
tq very much...^o^...are u a lecturer or something??
no, just waiting for the sumer semester to start up :)
i left out a bit there.... the ^(3/2)
2(2+tan^-1(3x))^(3/2) ------------------- +C ?? 9 \[\frac{2 (2+\tan^{-1}(3x))^{3/2}}{9} +C\]
sorry that was a wrong post for a different question

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