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

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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

$\blacksquare$


Sources