Question

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

Answer 1

\[\int\limits_{}^{}(2+\tan ^{-1}3x)^{1/2}\div(1+9x ^{2})\]

Answer 2

if u=2+tan^-1(3x)

du = 3x/(1+9x^2) dx

Answer 3

make that 3/(1+9x^2)

Answer 4

[S] u^(1/2)/3 du

Answer 5

2(2+tan^-1(3x))
--------------- +C ??
9

Answer 6

tq very much...^o^...are u a lecturer or something??

Answer 7

no, just waiting for the sumer semester to start up :)

Answer 8

i left out a bit there.... the ^(3/2)

Answer 9

2(2+tan^-1(3x))^(3/2)
------------------- +C ??
9

\[\frac{2 (2+\tan^{-1}(3x))^{3/2}}{9} +C\]

Answer 10

sorry that was a wrong post for a different question