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Calculate the integral
$\int{ \dfrac{ 1 }{ x ^{ 2 } -4 } }d{ x }$
$\dfrac { 1 } { 4 } \ln { \left( | \dfrac { 1 } { 2 } x - 1 | \right) } - \dfrac { 1 } { 4 } \ln { \left( | \dfrac { 1 } { 2 } x + 1 | \right) }$
Calculate the integral
$\displaystyle\int { \color{#FF6800}{ \dfrac { 1 } { x ^ { 2 } - 4 } } } d { \color{#FF6800}{ x } }$
 Substitute with $u = \dfrac { 1 } { \sqrt{ 4 } } x$ and calculate the integral 
$\left [ \color{#FF6800}{ \frac { 1 } { \sqrt{ 4 } } } \displaystyle\int { \color{#FF6800}{ \frac { 1 } { u ^ { 2 } - 1 } } } d { \color{#FF6800}{ u } } \right ] _ { \color{#FF6800}{ u } = \color{#FF6800}{ \frac { 1 } { \sqrt{ 4 } } } \color{#FF6800}{ x } }$
$\left [ \frac { 1 } { \sqrt{ 4 } } \displaystyle\int { \color{#FF6800}{ \frac { 1 } { u ^ { 2 } - 1 } } } d { \color{#FF6800}{ u } } \right ] _ { u = \frac { 1 } { \sqrt{ 4 } } x }$
 Calculate the integral using partial integration 
$\left [ \frac { 1 } { \sqrt{ 4 } } \left ( \color{#FF6800}{ \frac { 1 } { 2 } } \left [ \displaystyle\int { \color{#FF6800}{ v } ^ { \color{#FF6800}{ - } \color{#FF6800}{ 1 } } } d { \color{#FF6800}{ v } } \right ] _ { \color{#FF6800}{ v } = \color{#FF6800}{ u } \color{#FF6800}{ - } \color{#FF6800}{ 1 } } \color{#FF6800}{ - } \color{#FF6800}{ \frac { 1 } { 2 } } \left [ \displaystyle\int { \color{#FF6800}{ v } ^ { \color{#FF6800}{ - } \color{#FF6800}{ 1 } } } d { \color{#FF6800}{ v } } \right ] _ { \color{#FF6800}{ v } = \color{#FF6800}{ u } \color{#FF6800}{ + } \color{#FF6800}{ 1 } } \right ) \right ] _ { u = \frac { 1 } { \sqrt{ 4 } } x }$
$\left [ \frac { 1 } { \sqrt{ 4 } } \left ( \frac { 1 } { 2 } \left [ \displaystyle\int { \color{#FF6800}{ v } ^ { \color{#FF6800}{ - } \color{#FF6800}{ 1 } } } d { \color{#FF6800}{ v } } \right ] _ { v = u - 1 } - \frac { 1 } { 2 } \left [ \displaystyle\int { v ^ { - 1 } } d { v } \right ] _ { v = u + 1 } \right ) \right ] _ { u = \frac { 1 } { \sqrt{ 4 } } x }$
 Calculate the integral using the formula of $\int x^{-1} dx = \ln(|x|)$
$\left [ \frac { 1 } { \sqrt{ 4 } } \left ( \frac { 1 } { 2 } \left [ \ln { \left( | \color{#FF6800}{ v } | \right) } \right ] _ { v = u - 1 } - \frac { 1 } { 2 } \left [ \displaystyle\int { v ^ { - 1 } } d { v } \right ] _ { v = u + 1 } \right ) \right ] _ { u = \frac { 1 } { \sqrt{ 4 } } x }$
$\left [ \frac { 1 } { \sqrt{ 4 } } \left ( \frac { 1 } { 2 } \left [ \ln { \left( | \color{#FF6800}{ v } | \right) } \right ] _ { \color{#FF6800}{ v } = \color{#FF6800}{ u } \color{#FF6800}{ - } \color{#FF6800}{ 1 } } - \frac { 1 } { 2 } \left [ \displaystyle\int { v ^ { - 1 } } d { v } \right ] _ { v = u + 1 } \right ) \right ] _ { u = \frac { 1 } { \sqrt{ 4 } } x }$
 Return the substituted value 
$\left [ \frac { 1 } { \sqrt{ 4 } } \left ( \frac { 1 } { 2 } \ln { \left( | \color{#FF6800}{ u } \color{#FF6800}{ - } \color{#FF6800}{ 1 } | \right) } - \frac { 1 } { 2 } \left [ \displaystyle\int { v ^ { - 1 } } d { v } \right ] _ { v = u + 1 } \right ) \right ] _ { u = \frac { 1 } { \sqrt{ 4 } } x }$
$\left [ \frac { 1 } { \sqrt{ 4 } } \left ( \frac { 1 } { 2 } \ln { \left( | u - 1 | \right) } - \frac { 1 } { 2 } \left [ \displaystyle\int { \color{#FF6800}{ v } ^ { \color{#FF6800}{ - } \color{#FF6800}{ 1 } } } d { \color{#FF6800}{ v } } \right ] _ { v = u + 1 } \right ) \right ] _ { u = \frac { 1 } { \sqrt{ 4 } } x }$
 Calculate the integral using the formula of $\int x^{-1} dx = \ln(|x|)$
$\left [ \frac { 1 } { \sqrt{ 4 } } \left ( \frac { 1 } { 2 } \ln { \left( | u - 1 | \right) } - \frac { 1 } { 2 } \left [ \ln { \left( | \color{#FF6800}{ v } | \right) } \right ] _ { v = u + 1 } \right ) \right ] _ { u = \frac { 1 } { \sqrt{ 4 } } x }$
$\left [ \frac { 1 } { \sqrt{ 4 } } \left ( \frac { 1 } { 2 } \ln { \left( | u - 1 | \right) } - \frac { 1 } { 2 } \left [ \ln { \left( | \color{#FF6800}{ v } | \right) } \right ] _ { \color{#FF6800}{ v } = \color{#FF6800}{ u } \color{#FF6800}{ + } \color{#FF6800}{ 1 } } \right ) \right ] _ { u = \frac { 1 } { \sqrt{ 4 } } x }$
 Return the substituted value 
$\left [ \frac { 1 } { \sqrt{ 4 } } \left ( \frac { 1 } { 2 } \ln { \left( | u - 1 | \right) } - \frac { 1 } { 2 } \ln { \left( | \color{#FF6800}{ u } \color{#FF6800}{ + } \color{#FF6800}{ 1 } | \right) } \right ) \right ] _ { u = \frac { 1 } { \sqrt{ 4 } } x }$
$\left [ \color{#FF6800}{ \frac { 1 } { \sqrt{ 4 } } } \left ( \color{#FF6800}{ \frac { 1 } { 2 } } \ln { \left( | \color{#FF6800}{ u } \color{#FF6800}{ - } \color{#FF6800}{ 1 } | \right) } \color{#FF6800}{ - } \color{#FF6800}{ \frac { 1 } { 2 } } \ln { \left( | \color{#FF6800}{ u } \color{#FF6800}{ + } \color{#FF6800}{ 1 } | \right) } \right ) \right ] _ { \color{#FF6800}{ u } = \color{#FF6800}{ \frac { 1 } { \sqrt{ 4 } } } \color{#FF6800}{ x } }$
 Return the substituted value 
$\color{#FF6800}{ \dfrac { 1 } { \sqrt{ 4 } } } \left ( \color{#FF6800}{ \dfrac { 1 } { 2 } } \ln { \left( | \color{#FF6800}{ \dfrac { 1 } { \sqrt{ 4 } } } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } | \right) } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 2 } } \ln { \left( | \color{#FF6800}{ \dfrac { 1 } { \sqrt{ 4 } } } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } | \right) } \right )$
$\color{#FF6800}{ \dfrac { 1 } { \sqrt{ 4 } } } \left ( \color{#FF6800}{ \dfrac { 1 } { 2 } } \ln { \left( | \color{#FF6800}{ \dfrac { 1 } { \sqrt{ 4 } } } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } | \right) } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 2 } } \ln { \left( | \color{#FF6800}{ \dfrac { 1 } { \sqrt{ 4 } } } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } | \right) } \right )$
 Simplify the expression 
$\color{#FF6800}{ \dfrac { 1 } { 4 } } \ln { \left( | \color{#FF6800}{ \dfrac { 1 } { 2 } } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } | \right) } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 4 } } \ln { \left( | \color{#FF6800}{ \dfrac { 1 } { 2 } } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } | \right) }$
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