Question

I need help with this integral:
\[\frac{2}{4t^{2}-9}dt\]

Answer 1

\[I=\int\limits \frac{ 2 }{ 4t^2-9 }dt=\int\limits \frac{ 2 }{ \left( 2t+3 \right)\left( 2t-3 \right) }dt\]

Answer 2

I need to use this

Answer 3

\[\int\limits \frac{ 2 }{ 4t^2-9 }dt\]
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.

Answer 4

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

Answer 5

\[x=0, 2=A(-3)+B(3)\rightarrow \frac{2+3A}{3}=B\]
yeah i dont remember how to do this

Answer 6

\[=\int\limits \left[ \frac{ 2 }{ \left( 2t+3 \right)\left( -3-3 \right) }+\frac{ 2 }{ \left( 3+3 \right)\left( 2t-3 \right) } \right]dt\]
\[=-\int\limits \frac{ 2 }{ 6\left( 2t+3 \right) }dt+\int\limits \frac{ 2 }{ 6\left( 2t-3 \right) }dt\]
\[=-\frac{ 1 }{ 6 }\ln \left( 2t+3 \right)+\frac{ 1 }{ 6 }\ln \left( 2t-3 \right)+c\]
\[=\frac{ 1 }{ 6 }\left[ -\ln \left( 2t+3 \right)+\ln \left( 2t-3 \right) \right]+c\]
\[=\frac{ 1 }{ 6 }\ln \frac{ 2t-3 }{ 2t+3 }+c\]

Answer 7

Choose more clever values for x, not x=0
:)

How about x=3/2, what does that give you?

Answer 8

x=3/2\[2\rightarrow A(0)+B(6)\rightarrow B=\frac{1}{3}\]
x=-3/2\[2\rightarrow A(-6)+B(0)\rightarrow A=\frac{-1}{3}\]

Answer 9

Mmm looks good \c:/

Answer 10

\[\frac{-1}{3}\int\frac{1}{2t+3}+\frac{1}{3}\int\frac{1}{2t-3}\rightarrow\]
\[\frac{-1}{6}ln|2t+3|+\frac{1}{6}ln|2t-3|\rightarrow\]
\[\frac{1}{6}ln|\frac{2t-3}{2t+3}|\]

Answer 11

For some reason I just prefer trig substitution, but I guess partial fractions gets the same results too.

Answer 12

That's a little more hardcore @Kainui but works as well.