Mathematics

Evaluate 
$$\int \sqrt{1+t^{3}} dt$$


SOLUTION
We have,
$$I=\int _{  }^{  }{ \sqrt { 1+{ t^{ 3 } } }  } dt \\I ={ \int _{  }^{  }{ \left( { 1+{ t^{ 3 } } } \right)  } ^{ 1/2 } }dt \\I =\int _{  }^{  }{ \left( { 1+\dfrac { 1 }{ 2 } { t^{ 3 } }-\dfrac { 1 }{ 8 } { t^{ 6 } }....... } \right)  } dt \\I =\left( { t+\dfrac { 1 }{ 8 } { t^{ 4 } }-\dfrac { 1 }{ { 56 } } .{ t^{ 7 } }+...... } \right) +C $$

Hence, this is the answer.
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Subjective Medium Published on 17th 09, 2020
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