# Math Help - Harder Integration

1. ## Harder Integration

Thanks,

2. 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)

3. $\int_{0}^{12} \frac{\log_{4} (4+x)}{4+x}\cdot dx = \frac{1}{\ln 4}\cdot \int_{0}^{12} \frac{\ln (4+x)}{4+x}\cdot dx =$

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

Kind regards

$\chi$ $\sigma$

4. ## Thanks 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

5. Make the substitution a = ln(4+x), thus da = dx / (4+x)