Question

HELP! problem with partial fractions! I give medal!!

Answer 1

the integral that i'm having difficulties to solve by partial fractions is this one: \[\int\limits \frac{ -x^3+2x^2+8x+4 }{ x^4+4x^2}dx\]

Answer 2

What do I do after i get to this: \[\frac{ A }{ x }+\frac{ B }{ x^2 }+\frac{ Cx+D }{x^2+4 }\] ?????

Answer 3

You just have to sort through it!

That is right. Get all those denominators and put it all back together.

Answer 4

I1ve already done it tkhunny, but it's can't advance much...

Answer 5

A(x)(x^2 + 4) + B(x^2 + 4) + (Cx+D)(x^2) -- That's all.

Answer 6

\[\frac{-x^3+2x^2+8x+4}{x^2(x^2+4)}=\frac{ A }{ x }+\frac{ B }{ x^2 }+\frac{ Cx+D }{x^2+4 }\]
\[-x^3+2x^2+8x+4=Ax(x^2+4)+B(x^2+4)+(Cx+D)(x^2)\]
Expand, group common powers of \(x\), match up coefficients with left side, then solve the system.

Answer 7

I did it!! What I can't do is to find the values of A, B, C and D after this.

Answer 8

I'm stuck at this step... my mind just went blank

Answer 9

\[-x^3+2x^2+8x+4=Ax^3+4Ax+Bx^2+4B+Cx^3+Dx^2\]
yields the system
\[\begin{cases}A+C=-1\\B+D=2\\4A=8\\4B=4\end{cases}\]

Answer 10

A and V are pretty quick from that system.

Answer 11

I couldn't solv it because i forgot to put the x in 4Ax, and I wasn't noticing it. Thank you people!!!!

Answer 12

All that nice calculus and it falls apart on a little algebra? Let's get up to speed, shall we?