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

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Theorem

$\ds \int \frac {\d x} {\paren {p + q e^{a x} }^2} = \frac x {p^2} + \frac 1 {a p \paren {p + q e^{a x} } } - \frac 1 {a p^2} \ln \size {p + q e^{a x} } + C$


Proof

\(\ds z\) \(=\) \(\ds p + q e^{a x}\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d z} {\d x}\) \(=\) \(\ds a q e^{a x}\) Derivative of $e^{a x}$
\(\ds \) \(=\) \(\ds a \paren {z - p}\) in terms of $z$
\(\ds \leadsto \ \ \) \(\ds \int \frac {\d x} {p + q e^{a x} }\) \(=\) \(\ds \int \frac {\d z} {a \paren {z - p} z^2}\) Integration by Substitution
\(\ds \) \(=\) \(\ds \frac 1 a \int \frac {\d z} {z^2 \paren {z - p} }\) Primitive of Constant Multiple of Function
\(\ds \) \(=\) \(\ds \frac 1 a \paren {\frac {-1} {-p z} + \frac 1 {\paren {-p}^2} \ln \size {\frac {z - p} z} } + C\) Primitive of $\dfrac 1 {x^2 \paren {a x + b} }$
\(\ds \) \(=\) \(\ds \frac 1 {a p z} + \frac 1 {a p^2} \ln \size {\frac {z - p} z} + C\) simplifying
\(\ds \) \(=\) \(\ds \frac 1 {a p \paren {p + q e^{a x} } } + \frac 1 {a p^2} \ln \size {\frac {q e^{a x} } {p + q e^{a x} } } + C\) substituting for $z$
\(\ds \) \(=\) \(\ds \frac 1 {a p \paren {p + q e^{a x} } } + \frac 1 {a p^2} \ln \size {q e^{a x} } - \frac 1 {a p^2} \ln \size {p + q e^{a x} } + C\) Difference of Logarithms
\(\ds \) \(=\) \(\ds \frac 1 {a p \paren {p + q e^{a x} } } + \frac 1 {a p^2} \ln \size q + \frac 1 {a p^2} \ln \size {e^{a x} } - \frac 1 {a p^2} \ln \size {p + q e^{a x} } + C\) Sum of Logarithms
\(\ds \) \(=\) \(\ds \frac 1 {a p \paren {p + q e^{a x} } } + \frac 1 {a p^2} \ln \size q + \frac 1 {a p^2} \map \ln {e^{a x} } - \frac 1 {a p^2} \ln \size {p + q e^{a x} } + C\) $e^{a x}$ always positive
\(\ds \) \(=\) \(\ds \frac 1 {a p \paren {p + q e^{a x} } } + \frac 1 {a p^2} \map \ln {e^{a x} } - \frac 1 {a p^2} \ln \size {p + q e^{a x} } + C\) $\dfrac 1 {a p^2} \ln \size q$ subsumed into constant
\(\ds \) \(=\) \(\ds \frac 1 {a p \paren {p + q e^{a x} } } + \frac 1 {a p^2} a x - \frac 1 {a p^2} \ln \size {p + q e^{a x} } + C\) Exponential of Natural Logarithm
\(\ds \) \(=\) \(\ds \frac x {p^2} + \frac 1 {a p \paren {p + q e^{a x} } } - \frac 1 {a p^2} \ln \size {p + q e^{a x} } + C\) simplification

$\blacksquare$


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