∫ ( a x + b ) n d x = ( a x + b ) n + 1 a ( n + 1 ) + C (for n ≠ − 1 ) {\displaystyle \int (ax+b)^{n}dx={\frac {(ax+b)^{n+1}}{a(n+1)}}+C\qquad {\mbox{(for }}n\neq -1{\mbox{)}}\,\!} ∫ 1 a x + b d x = 1 a ln ⁡ | a x + b | + C {\displaystyle \int {\frac {1}{ax+b}}dx={\frac {1}{a}}\ln \left|ax+b\right|+C} More generally: ∫ 1 a x + b d x = { 1 a ln ⁡ | a x + b | + C − x < − b / a 1 a ln ⁡ | a x + b | + C + x > − b / a {\displaystyle \int {\frac {1}{ax+b}}\,dx={\begin{cases}{\frac {1}{a}}\ln \left|ax+b\right|+C^{-}&x<-b/a\\{\frac {1}{a}}\ln \left|ax+b\right|+C^{+}&x>-b/a\end{cases}}}