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.

`\frac1{y^2-16}=\frac18\left(\frac1{y-4}-\frac1{y+4}\right)`

Then

`\displaystyle\int(\mathrm dy)/(y^2-16)=\frac18\left(\int(\mathrm dy)/(y-4)+\int(\mathrm dy)/(y+4)\right)=\frac18\left(\ln|y-4|-\ln|y+4|\right)+C`

`=\frac18\ln\left|(y-4)/(y+4)\right|+C`

`=\ln\sqrt[8]\left+C`