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

## 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

$\blacksquare$