Question

integrate 1+2x^2/x^2(1+x^2)

Answer 1

SOLUTION

TO EVALUATE

$\displaystyle \sf{ \int \frac{1 + 2 {x}^{2} }{ {x}^{2}(1 + {x}^{2} ) } dx\: }$

EVALUATION

$\displaystyle \sf{ \int \frac{1 + 2 {x}^{2} }{ {x}^{2}(1 + {x}^{2} ) } dx\: }$

$= \displaystyle \sf{ \int \frac{(1 + {x}^{2} )+ {x}^{2} }{ {x}^{2}(1 + {x}^{2} ) } dx\: }$

$= \displaystyle \sf{ \int \frac{(1 + {x}^{2} ) }{ {x}^{2}(1 + {x}^{2} ) } dx\: + \int \frac{ {x}^{2} }{ {x}^{2}(1 + {x}^{2} ) } dx\:}$

$= \displaystyle \sf{ \int \frac{1 }{ {x}^{2} } dx\: + \int \frac{ 1 }{ 1 + {x}^{2} } dx\:}$

$= \displaystyle \sf{ \int { {x}^{ - 2} } dx\: + \int \frac{ 1 }{ 1 + {x}^{2} } dx\:}$

$= \displaystyle \sf{ \frac{ {x}^{ - 2 + 1} }{ - 2 + 1} + { \tan}^{ - 1}x } + c$

$= \displaystyle \sf{ \frac{ {x}^{ - 1} }{ - 1} + { \tan}^{ - 1}x } + c$

$= \displaystyle \sf{ - \frac{1}{x} + { \tan}^{ - 1}x } + c$

Where c is integration constant

FINAL ANSWER

$\displaystyle \sf{ \int \frac{1 + 2 {x}^{2} }{ {x}^{2}(1 + {x}^{2} ) } dx\: } = \displaystyle \sf{ - \frac{1}{x} + { \tan}^{ - 1}x } + c$

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