How to solve triple integral?

Question

$\int_{-2}^{5} \! \int_{-4}^{5} \! \int_{-4}^{3} \! \frac{1}{5x+8z+80} \, dx \, dy \, dz $

What to do with $\frac{1}{5x+8z+80}$ ? How to make it simplier?

Answer 1

You have

$$\int \dfrac{1}{ax+b}=\dfrac{\ln(ax+b)}{a}+C.$$

Then

$$\int_{-3}^4 \dfrac{dx}{5x+8z+80} = \left. \dfrac{\ln(5x+8z+80)}{5} \right |^{x=3}_{x=-4} = \dfrac{\ln(8z+105)}{5} -\dfrac{\ln(8z+60)}{5}. $$

Then you have

\begin{multline} \int_{-4}^5 \left(\dfrac{\ln(8z+105)}{5} -\dfrac{\ln(8z+60)}{5} \right)dy = \ln(8z+105) -\ln(8z+60) \\ - \dfrac{1}{4} \left( \dfrac{\ln(8z+105)}{5} -\dfrac{\ln(8z+60)}{5} \right). \end{multline}

The last integral is

$$ \int_{-2}^5 \left( \ln(8z+105) -\ln(8z+60)- \dfrac{1}{4} \left( \dfrac{\ln(8z+105)}{5} -\dfrac{\ln(8z+60)}{5} \right) \right)dz,$$

which is very ugly. Again, you have a 'simple form' of this integral

$$ \int \ln(ax+b) \, dx= \dfrac{(ax+b) \ln(ax+b)-ax}{a}. $$

For the first term

\begin{multline} \int_{-2}^5 \ln(8z+105) \,dz = \dfrac{(40+105)\ln(40+105)-40}{8} \\ - \dfrac{(-16+105)\ln(-16+105)+16}{8}. \end{multline}

And as you may know, the algebra will be tedious for the remaining three terms.