Integrating 6^2x, I did (6^2x)/(2ln(2x)) + C, but....

[johnboy]

My problem is to integrate 6^2x... so I just did (6^2x)/(2ln(2x)) + C, but my book says the answer is (2^2x-1*3^2x) / ln6. Is my answer acceptable? Thanks.

 

[pka]

Your answer is equivalent.
\(\displaystyle \L
\frac{{6^{2x} }}{{2\ln (6)}} = \frac{{6^{2x} }}{{2\left( {\ln (2) + \ln (3)} \right)}} = \frac{{36^x }}{{2\left( {\ln (2) + \ln (3)} \right)}}\)

 

[soroban]

Hello, johnboy!

My problem is: \(\displaystyle \L\,\int 6^{^{2x}}dx\) ... so I just did: \(\displaystyle \L\,\frac{6^{^{2x}}}{2\cdot\ln(2x)}\,+\,C\)

but my book says the answer is: \(\displaystyle \L\,\frac{2^{^{2x-1}}\cdot\,3^{^{2x}}}{\ln6}\)
Is my answer acceptable? . . . no

Did you type your answer wrong?

It should be: \(\displaystyle \L\,\frac{6^{^{2x}}}{2\cdot\ln6}\,+\,C\)
. . . . . . . . . . . . . ↑

Your book simpilified it further: \(\displaystyle \L\;\frac{6^{^{2x}}}{2\cdot\ln6}\;= \;\frac{2^{^{2x}}\cdot\,3^{^{2x}}}{2\cdot\ln6}\)
\(\displaystyle \;\)and cancelled a 2: \(\displaystyle \L\;\frac{2^{^{2x-1}}\cdot\,3^{^{2x}}}{\ln6}\)