Question

Integrate by parts or trig substition

integral or anti derivative of
dy/(y^2-16)

Answer 1

The simplest method probably would be to split up the integrand into partial fractions.

$\dfrac1{y^2-16}=\dfrac18\left(\dfrac1{y-4}-\dfrac1{y+4}\right)$

Then

$\displaystyle\int\frac{\mathrm dy}{y^2-16}=\frac18\left(\int\frac{\mathrm dy}{y-4}+\int\frac{\mathrm dy}{y+4}\right)=\frac18\left(\ln|y-4|-\ln|y+4|\right)+C$

$=\dfrac18\ln\left|\dfrac{y-4}{y+4}\right|+C$

$=\ln\sqrt[8]{\left|\dfrac{y-4}{y+4}\right|}+C$