Question

# Evaluate : $$\int \dfrac { x ^ { 2 } + 1 } { x ^ { 4 } + x ^ { 2 } + 1 } d x$$

Solution

## $$\int { \dfrac { { x }^{ 2 }+1 }{ { x }^{ 4 }+{ x }^{ 2 }+1 } } dx$$$$=\int { \dfrac { { x }^{ 2 }+1 }{ { x }^{ 4 }+{ 2x }^{ 2 }+1-{ x }^{ 2 } } } dx$$$$=\int { \dfrac { { x }^{ 2 }+1 }{ { \left( { x }^{ 2 }+1 \right) }^{ 2 }-{ x }^{ 2 } } } dx$$$$=\int { \dfrac { { x }^{ 2 }+1 }{ \left( { x }^{ 2 }+1-x \right) \left( { x }^{ 2 }+1+x \right) } } dx$$$$=\dfrac { 1 }{ 2 } \int { \left[ \dfrac { 1 }{ { x }^{ 2 }+1-x } +\dfrac { 1 }{ { x }^{ 2 }+1+x } \right] } dx$$$$=\dfrac { 1 }{ 2 } \int { \dfrac { 1 }{ { x }^{ 2 }-x+1 } dx } +\dfrac { 1 }{ 2 } \int { \dfrac { 1 }{ { x }^{ 2 }+x+1 } dx }$$$$=\dfrac { 1 }{ 2 } \int { \dfrac { 1 }{ { x }^{ 2 }-2.\dfrac { 1 }{ 2 } .x+\dfrac { 1 }{ 4 } +\dfrac { 3 }{ 4 } } dx } +\dfrac { 1 }{ 2 } \int { \dfrac { 1 }{ { x }^{ 2 }+2.\dfrac { 1 }{ 2 } x+\dfrac { 1 }{ 4 } +\dfrac { 3 }{ 4 } } dx }$$$$=\dfrac { 1 }{ 2 } \int { \dfrac { 1 }{ { \left( x-\dfrac { 1 }{ 2 } \right) }^{ 2 }+{ \left( \dfrac { \sqrt { 3 } }{ 2 } \right) }^{ 2 } } dx } +\dfrac { 1 }{ 2 } \int { \dfrac { 1 }{ { \left( x+\dfrac { 1 }{ x } \right) }^{ 2 }+\dfrac { 3 }{ 4 } } dx }$$$$=\dfrac { 1 }{ 2 } \times \dfrac { 2 }{ \sqrt { 3 } } { \tan }^{ -1 }\dfrac { x-1/2 }{ \sqrt { 3 } /2 } +\dfrac { 1 }{ 2 } \times \dfrac { 2 }{ \sqrt { 3 } } { \tan }^{ -1 }\dfrac { x+1/2 }{ \sqrt { 3 } /2 } +C$$$$=\dfrac { 1 }{ \sqrt { 3 } } { \tan }^{ -1 }\left( \dfrac { 2x-1 }{ \sqrt { 3 } } \right) +\dfrac { 1 }{ \sqrt { 3 } } { \tan }^{ -1 }\left( \dfrac { 2x+1 }{ \sqrt { 3 } } \right) +C$$Mathematics

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