Mathematics

If $$\displaystyle I=\int \log\left ( \sqrt{1-x}+\sqrt{1+x} \right )dx,$$ then I is equal to


ANSWER

$$\displaystyle x\log \left ( \sqrt{1-x}+\sqrt{1+x} \right )+\frac{1}{2}\sin^{-1}x-\frac{1}{2}x+C$$


SOLUTION
Let $$\displaystyle I=\int { \log { \left( \sqrt { 1-x } +\sqrt { 1+x }  \right)  }  } dx$$

$$\displaystyle =x\log { \left( \sqrt { 1-x } +\sqrt { 1+x }  \right)  } -\int { \frac { x }{ \left( \sqrt { 1-x } +\sqrt { 1+x }  \right)  }  } \left( \frac { -1 }{ 2\sqrt { 1-x }  } +\frac { 1 }{ 2\sqrt { 1+x }  }  \right) dx$$

$$\displaystyle =x\log { \left( \sqrt { 1-x } +\sqrt { 1+x }  \right)  } +\frac { 1 }{ 2 } { I }_{ 1 }$$

Where $$\displaystyle { I }_{ 1 }=\int { \frac { \sqrt { 1+x } -\sqrt { 1-x }  }{ \sqrt { 1+x } +\sqrt { 1-x }  }  } \frac { x }{ \sqrt { 1-{ x }^{ 2 } }  } dx$$

$$\displaystyle =\int { \frac { { \left( \sqrt { 1+x } -\sqrt { 1-x }  \right)  }^{ 2 } }{ 1+x-1+x }  } \frac { x }{ \sqrt { 1-{ x }^{ 2 } }  } dx$$

$$\displaystyle =\frac { 1 }{ 2 } \int { \frac { \\ 1+x+1-x-2\sqrt { 1-{ x }^{ 2 } }  }{ \sqrt { 1-{ x }^{ 2 } }  }  } dx=\int { \left( \frac { 1 }{ \sqrt { 1-{ x }^{ 2 } }  } -1 \right)  } dx=\sin ^{ -1 }{ x } -x$$

Therefore $$\displaystyle I=x\log { \left( \sqrt { 1-x } +\sqrt { 1+x }  \right)  } +\frac { 1 }{ 2 } \left( \sin ^{ -1 }{ x } -x \right) +c$$
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Single Correct Hard Published on 17th 09, 2020
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