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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?

jgon
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yara
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  • @Cœur No worries, sorry, everyone was editing all at once, I tried to get an edit that hopefully includes all the positive changes. – jgon Jan 02 '18 at 04:44
  • Why was this question downvoted again? – Crescendo Jan 02 '18 at 04:44
  • @yara does answer come out to be different using partial fraction? – prog_SAHIL Jan 02 '18 at 05:09
  • I'm not sure $\frac14\text{arctan...}$ is the right answer, it should be $-\frac14\text{arctanh...}$(unless I'm missing something) – ℋolo Jan 02 '18 at 05:09
  • @prog_SAHIL the answer will be the same, it is just that he got the wrong answer – ℋolo Jan 02 '18 at 05:13
  • this was the answer i found using the partial fraction method 1/8ln(|2x−3|)−1/8ln(|2x+1|)+C. – yara Jan 02 '18 at 05:17
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    I'll now explain how I got to $\text{-arctanh}$ using complete the square: $\frac1{4x^2-4x-3}=\frac1{(2x-1)^2-4}$ set $u=2x-1\implies du=2$ so we left with $\frac12\frac1{u^2-4}=-\frac12\frac1{2^2-u^2}$. The integral of $\frac1{a^2-x^2}=\frac{\text{arctanh}(x/a)}a$ hence the integral of $-\frac12\frac1{2^2-u^2}=-\frac14\text{arctanh}(u/2)=-\frac14\text{arctanh}(x-0.5)$ – ℋolo Jan 02 '18 at 05:21

2 Answers2

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So in a comment on a now deleted post, dxiv wrote $\int \frac{1}{1-x^2}\ne \arctan(x) = \int \frac{1}{1+x^2}$. This is what's going on here. The integral of $\int \frac{1}{q(x)}$ where $q$ is a quadratic polynomial depends on whether its roots are real or complex. If its roots are real, $q$ factors as real polynomials and you can use partial fraction decomposition. If its roots are complex, you can complete the square and use the arctangent method. Now that's the core issue. Credit to dxiv for pointing it out.

Side note, whether the roots are real or complex depends on the discriminant of $q$. In this case the discriminant is $4^2+4^2\cdot 3=64 > 0$, so the roots are real, so you need to use partial fraction decomposition.

Side side note, there's actually a third case when the discriminant is 0, and $q$ has a double root. In this case, you can change variables to make $q=x^2$, then $\int \frac{1}{x^2} = \frac{-1}{x}$.

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

haqnatural
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