Mathematics

If $$\int { \cfrac { xdx }{ \sqrt { 1+{ x }^{ 2 }+\sqrt { { \left( 1+{ x }^{ 2 } \right)  }^{ 3 } }  }  }  } =k\sqrt { 1+\sqrt { 1+{ x }^{ 2 } }  } +C$$ then k equals to


SOLUTION
Put, $$1+{ x }^{ 2 }={ z }^{ 2 }$$
$$2zdz=2xdx$$
$$\int { \cfrac { xdx }{ \sqrt { 1+{ x }^{ 2 }+\sqrt { { \left( 1+{ x }^{ 2 } \right)  }^{ 3 } }  }  }  } $$
$$=\int { \cfrac { zdz }{ \sqrt { { z }^{ 2 }+{ z }^{ 3 } }  }  } $$
$$=\int { \cfrac { zdz }{ z\sqrt { 1+{ z } }  }  } =\int { \cfrac { dz }{ \sqrt { 1+{ z } }  }  } =\int { { \left( 1+z \right)  }^{ -{ 1 }/{ 2 } }\cdot dz } $$
$$=\cfrac { 1 }{ -{ 1 }/{ 2 }+1 } \times { \left( 1+z \right)  }^{ -{ 1 }/{ 2 }+1 }+C$$
$$=\cfrac { 1 }{ 1/2 } { (1+z) }^{ { 1 }/{ 2 } }+C$$
$$2\sqrt { 1+z } +C$$
$$=2\sqrt { 1+\sqrt { 1+{ x }^{ 2 } }  } +C$$
Ans: k=2
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Subjective Medium Published on 17th 09, 2020
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