Mathematics

# Solve:$I = \int {\dfrac{{dx}}{{18 - 4x - {x^2}}}}$

##### SOLUTION
$\int \dfrac { d x } { 18 - 4 x - x ^ { 2 } }$
$= - 1 \int \dfrac { d x } { x ^ { 2 } + 4 x - 18 }$
$= - 1 \int \dfrac { d x } { ( x + 2 ) ^ { 2 } - ( \sqrt { 22 } ) ^ { 2 } }$
$= - 1 \dfrac { 1 } { 2 \cdot \sqrt { 22 } } \ln \left( \dfrac { x + 2 - \sqrt { 2 } 2 } { x + 2 + \sqrt { 2 } } \right) + C$
$= \dfrac { - 1 } { 2 \sqrt { 22 } } \ln ( \dfrac { x + 2 - \sqrt { 2 } 2 } { x + 2 + \sqrt { 2 } 2 } ) + C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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1 Verified Answer | Published on 17th 09, 2020

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