<s>[INST] Differences between using partial fractions or completing the square to solve an integral? I have this question 
$$\int \frac{1}{4x^2-4x-3}\, dx$$ 
I tried to solve it by using completing square method, and I got
$$\frac{1}{4} \arctan\left(\frac{2x-1}{2}\right),$$
but when I saw the answers, I found that it should be solved by partial fraction.
So what are the differences between these two methods?
 [/INST] $$\int { \frac { dx }{ 4x^{ 2 }-4x-3 }  } =\int { \frac { dx }{ \left( 2x+1 \right) \left( 2x-3 \right)  }  } =-\frac { 1 }{ 4 } \int { \left[ \frac { 1 }{ 2x+1 } -\frac { 1 }{ 2x-3 }  \right] dx } \\ =-\frac { 1 }{ 4 } \left[ \int { \frac { dx }{ 2x+1 }  } -\int { \frac { dx }{ 2x-1 }  }  \right] =-\frac { 1 }{ 8 } \left[ \int { \frac { d\left( 2x+1 \right)  }{ 2x+1 }  } -\int { \frac { d\left( 2x-1 \right)  }{ 2x-1 }  }  \right] =\\ =-\frac { 1 }{ 8 } \left[ \ln { \left| 2x+1 \right| -\ln { \left| 2x-1 \right|  }  }  \right] =-\frac { 1 }{ 8 } \ln { \left| \frac { 2x+1 }{ 2x-1 }  \right|  } +C=\\ \\ or\\ -\frac { 1 }{ 8 } \ln { \left| \frac { 2x+1 }{ 2x-1 }  \right|  } +C=-\frac { 1 }{ 4 } \tanh ^{ -1 }{ \left( \frac { 2x+1 }{ 2x-1 }  \right)  } =\frac { 1 }{ 4 } \tanh ^{ -1 }{ \left( \frac { 2x+1 }{ 1-2x }  \right)  } \\ \\ $$
Obviously,here should be $\tanh ^{ -1 }{ x } $ not $\tan ^{ -1 }{ x } $
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