Mathematics

Solve : $$\displaystyle \int \dfrac{x^4dx}{(1+x^2)^3}$$ ?


SOLUTION
Given the integral,
$$\int { \dfrac { { x }^{ 4 }dx }{ { ({ x }^{ 2 }+1) }^{ 3 } }  } $$
using partial fraction we get,
$$=\int { \left( \dfrac { 1 }{ { x }^{ 2 }+1 } -\dfrac { 2 }{ { ({ x }^{ 2 }+1) }^{ 2 } } +\dfrac { 1 }{ { ({ x }^{ 2 }+1) }^{ 3 } }  \right)  } \\ =\int { \dfrac { 1 }{ { x }^{ 2 }+1 }  } dx-2\int { \dfrac { 1 }{ { ({ x }^{ 2 }+1) }^{ 2 } }  } dx+\int { \dfrac { 1 }{ { ({ x }^{ 2 }+1) }^{ 3 } }  } dx$$
Here,
$$\int { \dfrac { 1 }{ { x }^{ 2 }+1 }  } dx\\ =\tan^{-1} { (x) } $$
For, 
$$\int { \dfrac { 1 }{ { ({ x }^{ 2 }+1) }^{ 2 } }  } dx$$
applying reduction formula we get,
$$=\dfrac { x }{ 2({ x }^{ 2 }+1) } +\dfrac { 1 }{ 2 } \int { \dfrac { 1 }{ { x }^{ 2 }+1 }  } dx\\ =\dfrac { \tan^{-1} { (x) }  }{ 2 } +\dfrac { x }{ 2({ x }^{ 2 }+1) } $$
Again for,
$$\int { \dfrac { 1 }{ { ({ x }^{ 2 }+1) }^{ 3 } }  } dx$$
applying reduction formula we get,
$$=\dfrac { x }{ 4{ ({ x }^{ 2 }+1) }^{ 2 } } +\dfrac { 3 }{ 4 } \int { \dfrac { 1 }{ { ({ x }^{ 2 }+1) }^{ 2 } }  } dx\\ =\dfrac { 3\tan^{-1} { (x) }  }{ 8 } +\dfrac { 3x }{ 8({ x }^{ 2 }+1) } +\dfrac { x }{ 4{ ({ x }^{ 2 }+1) }^{ 2 } } \\ \therefore \int { \dfrac { 1 }{ { x }^{ 2 }+1 }  } dx-2\int { \dfrac { 1 }{ { ({ x }^{ 2 }+1) }^{ 2 } }  } dx+\int { \dfrac { 1 }{ { ({ x }^{ 2 }+1) }^{ 3 } }  } dx\\ =\dfrac { 3\tan^{-1} { (x) }  }{ 8 } -\dfrac { 5x }{ 8({ x }^{ 2 }+1) } +\dfrac { x }{ 4{ ({ x }^{ 2 }+1) }^{ 2 } } $$
Hence,
$$\int { \dfrac { { x }^{ 4 }dx }{ { ({ x }^{ 2 }+1) }^{ 3 } }  } \\ =\dfrac { 3\tan^{-1} { (x) }  }{ 8 } -\dfrac { 5x }{ 8({ x }^{ 2 }+1) } +\dfrac { x }{ 4{ ({ x }^{ 2 }+1) }^{ 2 } } +C\\ =\dfrac { 3\tan^{-1} { (x) }  }{ 8 } +\dfrac { -5{ x }^{ 3 }-3x }{ 8{ ({ x }^{ 2 }+1) }^{ 2 } } +C.$$
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Subjective Medium Published on 17th 09, 2020
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