Question

While integrating by partial fractions, I end up with A(x+1)^2+B(2x)(x+1)+(Cx+D)(2x)=4x^2-1 ... No matter what I try, I end up with a no-solution system. Any help?

Answer 1

If i'm not mistaken, for X = 0, I can get A = -1 as a solution.

Answer 2

could you post the original problem?

Answer 3

Set x = -1:
\[2C-2D=3\]
Set x = 0:
\[A=-1\]
Set x = 1:
\[4A+4B+2C+2D=3\]
Set x = 2:
\[9A+12B+8C+4D=15\]Can you solve this system?

Answer 4

\[\int\limits(4x^2-1)\div2x(x^2+2x+1)\]

Answer 5

use x=-1 to find one equality, then look at the coefficients on the powers of x, for instance, for x^2 we have A + 2B + 2C = 4 => 2B + 2C = 5
for x we have 2A + 2B + 2D = 0 => B + D = 1, you now have 3 equalities and can solve for each variable

Answer 6

Online calculators, as well as pen&paper say no solution..

Answer 7

lies and slander, one sec while I type a solution.

Answer 8

\[\int \frac{4x^2-1}{2x(x^2+2x+1)}dx=\frac12\int\frac{4x^2-1}{x(x^2+2x+1)}dx=\frac12\int\frac{4x^2-1}{x(x+1)^2}dx\]
\[=\frac12\int\left(\frac{5}{x+1}-\frac{3}{(x+1)^2}-\frac1x\right)dx=-\frac12\int\frac1xdx-\frac32\int \frac{1}{(x+1)^2}dx+\frac52\int\frac1{x+1}dx\]
Let:
u=x+1
du=dx
\[=-\frac12\int\frac1xdx-\frac32\int\frac1{u^2}du+\frac52\int\frac1{u}du\]
\[=-\frac12\ln(x)+\frac3{2u}+\frac52\ln(u)+C\]
\[=\frac{-u\ln(x)+5u\ln(u)+3}{2u}+C\]
plug back in for u and simplify: you should find
\[=\frac{1}{2}\left(\frac3{x+1}-\ln(x)+5\ln(x+1)\right)+C\]

Answer 9

Uhh, I seen that on Wolfram Alpha, now how do I get to the part where there 5, 3 and -1 are numerators?

Answer 10

you set x=1 to make your system of inequalities... I'm not good at explaining partial fractions, the two above gave a decent shot at it... a bit more in depth explanation of that step is here:
http://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/partialfracdirectory/PartialFrac.html

Answer 11

I keep getting no solution systems...