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Question

$$\displaystyle \int { \frac { \sqrt { x }  }{ \sqrt { { x }^{ 3 }+4 }  } dx } $$ is equal to


A
23ln(2x3x34)+c
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B
23ln(2x3+x34)+c
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C
23ln(2x3x34)+c
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D
None of these
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Solution

The correct option is B $$\displaystyle \frac { 2 }{ 3 } \ln { \left( \frac { 2 }{ \sqrt { { x }^{ 3 } } -\sqrt { { x }^{ 3 }-4 }  }  \right)  } +c$$
Given: $$\displaystyle \int { \frac { \sqrt { x }  }{ \sqrt { { x }^{ 3 }+4 }  } dx } $$

Substitute $$\displaystyle { x }^{\tfrac 32 }=2\tan { \theta  } \Rightarrow \frac { 3 }{ 2 } { x }^{\tfrac 12 }dx=2\sec ^{ 2 }{ \theta  } d\theta $$

$$\displaystyle \therefore I=\int { \dfrac { \dfrac { 4 }{ 3 } \sec ^{ 2 }{ \theta  } d\theta  }{ \sqrt { 4\tan ^{ 2 }{ \theta  } +4 }  }  } =\frac { 2 }{ 3 } \int { \sec { \theta  } d\theta  } $$

$$\displaystyle =\frac { 2 }{ 3 } \ln { \left( \sec { \theta  } +\tan { \theta  }  \right)  } +c=\frac { 2 }{ 3 } \ln { \left( \sqrt { \frac { { x }^{ 3 }-4 }{ 4 }  } +\frac { { x }^{ \frac{3}{2} } }{ 2 }  \right)  } +c$$

$$\displaystyle =\frac { 2 }{ 3 } \ln { \left( \frac { \sqrt { { x }^{ 3 } } +\sqrt { { x }^{ 3 }-4 }  }{ 2 }  \right)  } +c=\frac { 2 }{ 3 } \ln { \left( \frac { { x }^{ 3 }-\left( { x }^{ 3 }-4 \right)  }{ 2\left( \sqrt { { x }^{ 3 } } +\sqrt { { x }^{ 3 }-4 }  \right)  }  \right)  } +c$$

$$\displaystyle =\frac { 2 }{ 3 } \ln { \left( \frac { 2 }{ \sqrt { { x }^{ 3 } } -\sqrt { { x }^{ 3 }-4 }  }  \right)  } +c$$

Mathematics

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