Primitive of Reciprocal of p plus q by Exponential of a x

Theorem

$\displaystyle \int \frac {\d x} {p + q e^{a x} } = \frac x p - \frac 1 {a p} \ln \size {p + q e^{a x} } + C$

Proof

 $\displaystyle z$ $=$ $\displaystyle p + q e^{a x}$ $\displaystyle \leadsto \ \$ $\displaystyle \frac {\d z} {\d x}$ $=$ $\displaystyle a q e^{a x}$ Derivative of $e^{a x}$ $\displaystyle$ $=$ $\displaystyle a \paren {z - p}$ in terms of $z$ $\displaystyle \leadsto \ \$ $\displaystyle \int \frac {\d x} {p + q e^{a x} }$ $=$ $\displaystyle \int \frac {\d z} {a \paren {z - p} z}$ Integration by Substitution $\displaystyle$ $=$ $\displaystyle \frac 1 a \int \frac {\d z} {z \paren {z - p} }$ Primitive of Constant Multiple of Function $\displaystyle$ $=$ $\displaystyle \frac 1 a \paren {\frac 1 {-p} \ln \size {\frac z {z - p} } } + C$ Primitive of $\dfrac 1 {x \paren {a x + b} }$ $\displaystyle$ $=$ $\displaystyle \frac {-1} {a p} \ln \size {\frac {p + q e^{a x} } {q e^{a x} } } + C$ substituting for $z$ $\displaystyle$ $=$ $\displaystyle \frac {-1} {a p} \paren {\ln \size {p + q e^{a x} } - \ln \size {q e^{a x} } } + C$ Difference of Logarithms $\displaystyle$ $=$ $\displaystyle \frac {-1} {a p} \paren {\ln \size {p + q e^{a x} } - \paren {\ln \size {e^{a x} } + \ln \size q} } + C$ Sum of Logarithms $\displaystyle$ $=$ $\displaystyle \frac {-1} {a p} \ln \size {p + q e^{a x} } + \frac 1 {a p} \map \ln {e^{a x} } + \frac 1 {a p} \ln \size q + C$ $e^{a x}$ always positive $\displaystyle$ $=$ $\displaystyle \frac {-1} {a p} \ln \size {p + q e^{a x} } + \frac 1 {a p} \map \ln {e^{a x} } + C$ $\dfrac 1 {a p} \ln \size q$ subsumed into arbitrary constant $\displaystyle$ $=$ $\displaystyle \frac {-1} {a p} \ln \size {p + q e^{a x} } + \frac 1 {a p} a x + C$ Exponential of Natural Logarithm $\displaystyle$ $=$ $\displaystyle \frac x p - \frac 1 {a p} \ln \size {p + q e^{a x} } + C$ simplification

$\blacksquare$