Question

another integral
\[\int \frac{1}{y^{2}+y} dy\]

Answer 1

sorry I feel like I should know this

Answer 2

partial fractions.... find A and B in the decomposition for the integrand:
\[\huge \frac{1}{y^2+y}=\frac{A}{y}+\frac{B}{y+1} \]

Answer 3

ok

Answer 4

did you do this before?

Answer 5

I believe so, lets see if I remember

Answer 6

you'll end up with:
\[\huge A(y+1)+By=1 \]

Answer 7

\[\huge \frac{A(y+1)y}{y}+\frac{B(y+1)y}{y+1}= 1\]

Answer 8

yes....

Answer 9

now all you gotta do is to choose "nice" values of y to make it easy in finding values for A and B...

for example, y=-1... solve for B... B=???

Answer 10

\[\huge A(y+1)+By=1\]
-1

Answer 11

B=-1 you mean right?

Answer 12

yes

Answer 13

ok...
so what other "nice" value for y will make it easy to find A?

Answer 14

0

Answer 15

yes... so A=??

Answer 16

1

Answer 17

there is also another way to do it without guessing for y values right?

Answer 18

good...
so now you have:
\[\huge \frac{1}{y^2+y}=\frac{1}{y}-\frac{1}{y+1} \]

Answer 19

yes, thanks!

Answer 20

hmm... they weren't actually guesses... they were "convenient" values for y...

Answer 21

I know, but I thought there was a way.
A+B=...
Oh wait I think I remember... one sec

Answer 22

oh... i think i know what you're getting at.....

Answer 23

You can either do test cases or solve a system of equations or some of both methods. :-)

Answer 24

and yes....
if that denominator was something more complicated, you'd have to solve a system...

Answer 25

@dpaInc was doing test cases as much as I can tell from skimming

Answer 26

\[\huge A(y+1)+By=1\]

how would I do that again?

Answer 27

"convenient" values for y.. if y=-1, A will dissapear... if y=0, B will disappear...

Answer 28

I guess the problem isn't complicated enough to solve a system

Answer 29

I understand how we solved this problem, by eliminating A to solve for B and vice versa

Answer 30

no.... but even when doing the system, if i remember correctly you'd still have to do this "convenient" thingy...

Answer 31

makes sense

Answer 32

thanks again @dpaInc !

Answer 33

yw... seems like you've gone through all this before... are you reviewing?

Answer 34

yes

Answer 35

nice work....:)

Answer 36

I guess I have forgotten almost everything about solving integrals

Answer 37

they'll come back to you....:)

Answer 38

I'm patiently waiting for that to happen LOL! thanks :)

Answer 39

:)