Kkutie7
  • Kkutie7
I need help with this integral: \[\frac{2}{4t^{2}-9}dt\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[I=\int\limits \frac{ 2 }{ 4t^2-9 }dt=\int\limits \frac{ 2 }{ \left( 2t+3 \right)\left( 2t-3 \right) }dt\]
Kkutie7
  • Kkutie7
I need to use this
Owlcoffee
  • Owlcoffee
\[\int\limits \frac{ 2 }{ 4t^2-9 }dt\] As appreciable, this is an integral solvabe by what is called "simple fractions method", you see, any fraction can be expressed as the sum of other two, take for example: \[\frac{ 3 }{ x.y } \iff \frac{ A }{ x }+\frac{ B }{ y }\] Where A and B will represent two different real numbers in such a way that if we effectuate the sum we wil obtain the the initial fraction. For this case, we wil work with the function inside: \[\frac{ 2 }{ 4t^2-9 }\] of course, in order for this to be applicabe, the denominator MUST be factorizable, which you must know by this point so I'll do it quickly: \[\frac{ 2 }{ 4t^2-9 } \iff \frac{ 2 }{ (2t+3)(2t-3) }\] And folowing up with the first explained property, we can rewrite it as: \[\frac{ 2 }{ (2t+3)(2t-3) } \iff \frac{ A }{ 2t+3 }+\frac{ B }{ 2t-3 }\] Now it's just a matter of finding the values for A and B.

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Kkutie7
  • Kkutie7
right i forgot about partial fractions. umm does it end up something like \[\frac{2}{(2t+3)(2t-3)}\rightarrow A(2t-3)+B(2t+3)\] I haven't done this in a long time
Kkutie7
  • Kkutie7
\[x=0, 2=A(-3)+B(3)\rightarrow \frac{2+3A}{3}=B\] yeah i dont remember how to do this
anonymous
  • anonymous
\[=\int\limits \left[ \frac{ 2 }{ \left( 2t+3 \right)\left( -3-3 \right) }+\frac{ 2 }{ \left( 3+3 \right)\left( 2t-3 \right) } \right]dt\] \[=-\int\limits \frac{ 2 }{ 6\left( 2t+3 \right) }dt+\int\limits \frac{ 2 }{ 6\left( 2t-3 \right) }dt\] \[=-\frac{ 1 }{ 6 }\ln \left( 2t+3 \right)+\frac{ 1 }{ 6 }\ln \left( 2t-3 \right)+c\] \[=\frac{ 1 }{ 6 }\left[ -\ln \left( 2t+3 \right)+\ln \left( 2t-3 \right) \right]+c\] \[=\frac{ 1 }{ 6 }\ln \frac{ 2t-3 }{ 2t+3 }+c\]
zepdrix
  • zepdrix
Choose more clever values for x, not x=0 :) How about x=3/2, what does that give you?
Kkutie7
  • Kkutie7
x=3/2\[2\rightarrow A(0)+B(6)\rightarrow B=\frac{1}{3}\] x=-3/2\[2\rightarrow A(-6)+B(0)\rightarrow A=\frac{-1}{3}\]
zepdrix
  • zepdrix
Mmm looks good \c:/
Kkutie7
  • Kkutie7
\[\frac{-1}{3}\int\frac{1}{2t+3}+\frac{1}{3}\int\frac{1}{2t-3}\rightarrow\] \[\frac{-1}{6}ln|2t+3|+\frac{1}{6}ln|2t-3|\rightarrow\] \[\frac{1}{6}ln|\frac{2t-3}{2t+3}|\]
Kainui
  • Kainui
For some reason I just prefer trig substitution, but I guess partial fractions gets the same results too.
Owlcoffee
  • Owlcoffee
That's a little more hardcore @Kainui but works as well.

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