Question

Please make sure I'm doing this right.
\[\int\frac{3y^{3}+5y-1}{y^{3}+y}dy\]

Answer 1

\[\rightarrow \int\frac{3y^{3}+5y-1}{y(y^{2}-1)}dy\]

Answer 2

\[\rightarrow\int\frac{3y^{3}+5y-1}{y(y-1)(y+1)}dy\]
\[\rightarrow \frac{3y^{3}+5y-1}{y(y-1)(y+1)}=\frac{A}{y}+\frac{B}{y-1}+\frac{C}{y+1}\]

Answer 3

\[\rightarrow 3y^{3}+5y-1=A(y-1)(y+1)+B(y)(y+1)+C(y)(y-1)\]

Answer 4

Using partial fraction decomposition for this program will be messy.

I haven't worked out all the details but here's what I have so far... I'm pretty sure all of my steps are legal... just breaking up the numerator more and more until we get rid of y^3 altogether.

Answer 5

\[\rightarrow x=0\rightarrow -1=-A, A=1\]
\[\rightarrow x=1,3+5-1=2B, B=\frac{7}{2}\]
\[\rightarrow x=-1,2C, C=\frac{-1}{2} \]
I don't see how things are messy, unless I'm doing it wrong

Answer 6

\[\rightarrow \int\frac{1}{y}+\frac{7}{2}\int\frac{1}{y-1}+\frac{-1}{2}\int\frac{1}{y+1}\]
\

Answer 7

\[\rightarrow ln|y|+\frac{7}{2}ln|y-1|-\frac{1}{2}ln|y+1|+C\]
is this right?

Answer 8

A = 1 ... agree

B = 7/2 ... agree

C = -1/2 ... disagree

Answer 9

C=-9/2?

Answer 10

nope

Answer 11

damn ok hold on

Answer 12

It is all wrong. You must do long division first and make the degree of numerator less than the degree of denominator..

Answer 13

@ganeshie8 no it's not all wrong. The only mistake I see is that C is the only thing that's off. Everything else looks fine

Answer 14

so: