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Mathematics 13 Online
OpenStudy (anonymous):

integral dx/xln3x

OpenStudy (anonymous):

yes

OpenStudy (rogue):

Substitute u = ln 3x.

OpenStudy (rogue):

\[u = \ln 3x, du = 3dx/x, \frac {1}{3} du = \frac {dx}{x}\]\[\int\limits \frac {dx}{x \ln 3x} \rightarrow \frac {1}{3} \int\limits \frac {du}{u} = \frac {1}{3} \ln \left| u \right| + C \rightarrow \frac {1}{3} \ln \left| \ln 3x \right|\]

OpenStudy (rogue):

Ah, right. That makes sense, I made a mistake.\[du= 3dx/3 x = dx/x\]

OpenStudy (rogue):

So its just those integrals, without the 1/3.

OpenStudy (anonymous):

i got \[\ln ^{2}3x \div2\]

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