Question

question

Answer 1

\[\int\limits (\frac{ 5 }{ (2x+1)(x-2) })dx \]

first step I did was this

\[\frac{ 5 }{ (2x+1)(x-2) } = \frac{ A }{ (2x+1) }+\frac{ B }{ (x-2) }\]

\[A(x-2)+B(2x+1) = 5 \]

\[Ax-2A+2Bx+B = 5+0x\]

Answer 2

\[(Ax-2Bx = 0) ; (2A+B = 5)\]

Answer 3

did I equate the coefficients correctly?

Answer 4

@mathmate

Answer 5

\[A-2B = 0 => A = 2B\]

\[2(2B)+B = 5 => 4B+B = 5B = 5 => B = 1 \]

A = 2
B = 1

2-(2(a)) = 0

Answer 6

\[\int\limits \frac{ 2 }{ (2x+1) }dx + \int\limits \frac{ 1 }{ (x-2) }dx\]

\[u = 2x+1 => \frac{ du }{ dx } = 2 => 2~dx = du => \frac{ du }{ 2 } = dx \]

\[\int\limits u~du*(\frac{ 1 }{ 2 })\]

\[2\ln|2x+1|+\ln|x-2|+C\]

Answer 7

sorry for bothering you I think I ended up with the answer I grouped the terms wrong

Answer 8

I get A+2B=0, B-2A=5
so A=-2,B=1

Answer 9

and so
\(\Large \int \frac{5}{(2x+1)(x-2)}=\int \frac{1}{x-2}-\frac{2}{2x+1}=ln(x-2)-ln(2x+1)\)

Answer 10

i guess I grouped the terms wrong

\[Ax-2A+2Bx+B = 5\]

Answer 11

it's simple but I miss this occasionally

so it's +Ax, +2bx, -2A, +B

Ax+2Bx = 0

B-2A = 5

Answer 12

was looking at what you did for term grouping

Answer 13

Yep, exactly.
It's easier on paper, because I circle all terms of the same degree. That way we don't miss easily.