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?
Best Answer
$$\int { \frac { dx }{ 4x^{ 2 }-4x-3 } } =\int { \frac { dx }{ \left( 2x+1 \right) \left( 2x-3 \right) } } =-\frac { 1 }{ 4 } \int { \left[ \frac { 1 }{ 2x+1 } -\frac { 1 }{ 2x-3 } \right] dx } \\ =-\frac { 1 }{ 4 } \left[ \int { \frac { dx }{ 2x+1 } } -\int { \frac { dx }{ 2x-1 } } \right] =-\frac { 1 }{ 8 } \left[ \int { \frac { d\left( 2x+1 \right) }{ 2x+1 } } -\int { \frac { d\left( 2x-1 \right) }{ 2x-1 } } \right] =\\ =-\frac { 1 }{ 8 } \left[ \ln { \left| 2x+1 \right| -\ln { \left| 2x-1 \right| } } \right] =-\frac { 1 }{ 8 } \ln { \left| \frac { 2x+1 }{ 2x-1 } \right| } +C=\\ \\ or\\ -\frac { 1 }{ 8 } \ln { \left| \frac { 2x+1 }{ 2x-1 } \right| } +C=-\frac { 1 }{ 4 } \tanh ^{ -1 }{ \left( \frac { 2x+1 }{ 2x-1 } \right) } =\frac { 1 }{ 4 } \tanh ^{ -1 }{ \left( \frac { 2x+1 }{ 1-2x } \right) } \\ \\ $$ Obviously,here should be $\tanh ^{ -1 }{ x } $ not $\tan ^{ -1 }{ x } $