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- May 7th 2009, 02:43 AMxibeleliHarder Integration
- May 7th 2009, 03:37 AMCalculus26
a )

Recall

log4(x) = ln(x)/ln(4)

b) Differentiate the Rhs

Keys tanh(x) ' = sech^2(x)

sinh(x)cosh(x) = 1/2sinh(2x) - May 7th 2009, 03:48 AMchisigma
$\displaystyle \int_{0}^{12} \frac{\log_{4} (4+x)}{4+x}\cdot dx = \frac{1}{\ln 4}\cdot \int_{0}^{12} \frac{\ln (4+x)}{4+x}\cdot dx =$

$\displaystyle = \frac{1}{2\cdot \ln 4}\cdot |\ln^{2}(4+x)|_{0}^{12} = \frac{3}{2}\cdot \ln 4$

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - May 9th 2009, 12:15 AMxibeleliThanks but still don't understand
Hi, thanks for your support guys but i still dont understand how u integration from step 2 to step 3 :(

- May 9th 2009, 12:26 PMjames
Make the substitution a = ln(4+x), thus da = dx / (4+x)