How do you integrate (1)/(e^x+2)? using partial fractions (another method is OK). Thank you!

Question

How do you integrate
`(1)/(e^x+2)`? using partial fractions (another method is OK). Thank you!

Answer 1

By substituting
`y=e^x+2`, or
`x=\ln(y-2)` so that
`\mathrm dx=(\mathrm dy)/(y-2)`, we have

`\displaystyle\int(\mathrm dx)/(e^x+2)=\int(\mathrm dy)/(y(y-2))`

Split the integrand into partial fractions:

`\frac1{y(y-2)}=\frac12\left(\frac1{y-2}-\frac1y\right)`

Then integrating gives

`\displaystyle\int(\mathrm dy)/(y(y-2))=\frac12(\ln|y-2|-\ln|y|)+C`

Back-substitute to get

`\displaystyle\int(\mathrm dx)/(e^x+2)=\frac12(\ln|e^x|-\ln|e^x+2|)+C=\frac{x-\ln(e^x+2)}2+C`