Mathematics

$$I = \int \frac { 1 } { \sqrt { 2 x ^ { 2 } + 3 x + 8 } } d x$$


SOLUTION
$$I=\int { \frac { 1 }{ \sqrt { 2{ x }^{ 2 }+3x+8 }  } dx } \\ =\frac { 1 }{ \sqrt { 2 }  } \int { \frac { 1 }{ \sqrt { { x }^{ 2 }+\frac { 3 }{ 2 } x+4 }  } dx } \\ =\frac { 1 }{ \sqrt { 2 }  } \int { \frac { 1 }{ \sqrt { { x }^{ 2 }+2\times \frac { 3 }{ 4 } x+\frac { 9 }{ 16 } +4-\frac { 9 }{ 16 }  }  } dx } \\ =\frac { 1 }{ \sqrt { 2 }  } \int { \frac { 1 }{ \sqrt { { \left( x+\frac { 3 }{ 4 }  \right)  }^{ 2 }+\frac { 55 }{ 16 }  }  } dx } \\ =\frac { 1 }{ \sqrt { 2 }  } \int { \frac { 1 }{ \sqrt { { \left( x+\frac { 3 }{ 4 }  \right)  }^{ 2 }+{ \left( \frac { \sqrt { 55 }  }{ 4 }  \right)  }^{ 2 } }  } dx } $$
let $$x+\frac { 3 }{ 4 } =u$$
$$dx=du$$
so $$I=\frac { 1 }{ \sqrt { 2 }  } \int { \frac { 1 }{ \sqrt { { u }^{ 2 }+{ \left( \frac { \sqrt { 55 }  }{ 4 }  \right)  }^{ 2 } }  } du } \\ =\frac { 1 }{ \sqrt { 2 }  } \ln { \left[ u+\sqrt { { u }^{ 2 }+{ \left( \frac { \sqrt { 55 }  }{ 4 }  \right)  }^{ 2 } }  \right]  } +c$$
putting $$u=x+\frac { 3 }{ 4 } $$
$$I=\frac { 1 }{ \sqrt { 2 }  } \ln { \left[ x+\frac { 3 }{ 4 } +\sqrt { { \left( x+\frac { 3 }{ 4 }  \right)  }^{ 2 }+{ \left( \frac { \sqrt { 55 }  }{ 4 }  \right)  }^{ 2 } }  \right]  } +c\\ =\frac { 1 }{ \sqrt { 2 }  } \ln { \left[ x+\frac { 3 }{ 4 } +\sqrt { { x }^{ 2 }+\frac { 3 }{ 2 } x+\frac { 9 }{ 16 } +\frac { 55 }{ 16 }  }  \right]  } +c\\ =\frac { 1 }{ \sqrt { 2 }  } \ln { \left[ x+\frac { 3 }{ 4 } +\sqrt { { x }^{ 2 }+\frac { 3 }{ 2 } x+4 }  \right]  } +c$$
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