Question

integrate (5x^2+20x+6) / (x^3+2x^2+x). show steps.

Answer 1

I would just factor first than use partial fractions

Answer 2

then*

Answer 3

this requires the "harder" method of partial fractions

Answer 4

yes i would do that too, and i would cheat to do it

Answer 5

because computing these partial fractions is going to be a bit of a pain since the denominator is \(x(x+1)^2\)
fortunately it is cooked up to give you integer answers

Answer 6

@eliassaab has a snap method for doing these but i don't know it

Answer 7

A = 6
5 = A + B + C
20 = 2A + B +C
What am i suppose to do with that? I can't seem to solve for B or C.

Answer 8

well if you know A = 6 then you have two equations with two unknowns

Answer 9

ya...

Answer 10

i dont get it

Answer 11

i need a value for B and C

Answer 12

hold on one second

Answer 13

ok

Answer 14

assuming your equations are right, the first one becomes \[5 = 6 + B + C\] or
\[-1=A+C\]

Answer 15

ok something is messed up let me do it with pencil

Answer 16

is there a good digital pen/pad i could buy? i wish i could show you my whole paper

Answer 17

i don't know of one
first off, lets get the answer

Answer 18

it is \[\frac{6}{x}-\frac{1}{x+1}+\frac{9}{(x+1)^2}\]

Answer 19

http://www.wolframalpha.com/input/?i=integrate+%285x%5E2%2B20x%2B6%29+%2F+%28x%5E3%2B2x%5E2%2Bx%29

Answer 20

now we have
\[\frac{5x^2+20x+6}{x(x+1)^2}=\frac{A}{x}+\frac{B}{x+1}+\frac{Cx+D}{(x+1)^2}\]

Answer 21

oh wait, i didnt do it exactly like that

Answer 22

here are the partial fractions, you don't have to ask wolfram to integrate
http://www.wolframalpha.com/input/?i=%285x^2%2B20x%2B6%29+%2F+%28x^3%2B2x^2%2Bx%29
it does it for you

Answer 23

i have x(x+1)(x+1) in denominator

Answer 24

http://www.wolframalpha.com/inputwe get \[a(x+1)^2+bx(x+1)+(cx+d)x=5x^2+20x+6\]

Answer 25

replacing \(x=0\) you get \(a=6\) right away

Answer 26

i have A/x + B/x+1 + C/x+1

Answer 27

that is a mistake

Answer 28

you have a repeated factor in the denominator

Answer 29

so it should be what i wrote above

Answer 30

\[\frac{5x^2+20x+6}{x(x+1)^2}=\frac{A}{x}+\frac{B}{x+1}+\frac{Cx+D}{(x+1)^2}\]

Answer 31

if you have a repeated factor, you need to take successive powers

Answer 32

however, i too made a mistake
since they are repeated it should be
\[\frac{5x^2+20x+6}{x(x+1)^2}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{(x+1)^2}\]

Answer 33

this gives \[a(x+1)^2+bx(x+1)+cx=5x^2+20x+6\] and as before replacing \(x\) by 0 gives \(a=6\)

Answer 34

then replacing
\(x\) by -1 gives \(-c=5-20+6)\) and so \(c=9\)

Answer 35

i have a monstrosity in my numerator

Answer 36

and since if you expand on the right you will get the coefficients of the \(x^2\) term as \(a+b\) and as \(a=6\) this is \(6+b\) you know \(6+b=5\) and so \(b=-1\)

Answer 37

don't expand

Answer 38

it is too much work

Answer 39

ok, then A(x+1)^3

Answer 40

numerator is
\[a(x+1)^2+bx(x+1)+cx=5x^2+20x+6\] there is no cube term

Answer 41

lets go slow

Answer 42

start with
\[\frac{5x^2+20x+6}{x(x+1)^2}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{(x+1)^2}\][

Answer 43

since \(x+1\) is a repeated factor

Answer 44

then if you add the stuff on the right, the numerator will be \[a(x+1)^2+bx(x+1)+cx=5x^2+20x+6\]

Answer 45

k, u lost me with that step...

Answer 46

or if you prefer
\[A(x+1)^2+Bx(x+1)+Cx=5x^2+20x+6\]
are these steps clear?

Answer 47

oh ok i thought maybe so

Answer 48

i thought A gets the denominators it doesn't have

Answer 49

lets go back to
\[\frac{5x^2+20x+6}{x(x+1)^2}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{(x+1)^2}\]

Answer 50

which would be x+1 and (x+1)^2

Answer 51

i see the problem

Answer 52

k, ya, i agree with that step

Answer 53

the denominator of your original thing is \(x(x+1)^2\), that is the least common multiple of the denominators on the right hand side as well

Answer 54

so if you were going to add
\[\frac{A}{x}+\frac{B}{x+1}+\frac{C}{(x+1)^2}\] then you would use your original denominator (that is the point of this)

Answer 55

since the lcd is \(x(x+1)^2\) and you have the term \(\frac{A}{x}\) you only need to multiply it top and bottom by \((x+1)^2\)

Answer 56

in other words when you add, that term would look like
\[\frac{A(x+1)^2}{x(x+1)^2}\]

Answer 57

so, in the future can i just look at the left denom and ask myself what's missing

Answer 58

oh yes exaclty!

Answer 59

just like you would with numbers in fact

Answer 60

assuming i didnt screw up that denom

Answer 61

right

Answer 62

cause if it's not in LCD it wont work i assume

Answer 63

so with that in mind, hopefully this line
\[A(x+1)^2+Bx(x+1)+Cx=5x^2+20x+6\] makes sense

Answer 64

you are thinking too hard i believe
you started with the denominator, that is what you have to end up with
they have to be the same

Answer 65

so if we get to the point where
\[A(x+1)^2+Bx(x+1)+Cx=5x^2+20x+6\] makes sense, then finding A, B and C is not that hard

Answer 66

ok, ya, i'm with you now

Answer 67

don't expand! it is a pain to do it that way

Answer 68

lemme see if i can continue

Answer 69

you can of course if you like
you can expand the left hand side, then equate like coefficients, and then solve a 3 by 3 system of equations, but that is really the long way to do it
substitute appropriate values for \(x\) to get what you want

Answer 70

what's the better way to do it

Answer 71

i usually expand the right side...

Answer 72

replace \(x=0\) solve for A in one step with your eyes

Answer 73

is that clear?

Answer 74

ok, yes, A=6

Answer 75

now what

Answer 76

replace \(x\) by what will give you zero in every other slot

Answer 77

i.e.\(-1\)
this takes only a tiny bit of arithmetic, although it is not as snappy as finding A

Answer 78

is that step clear?

Answer 79

no, can u give me the equation for B

Answer 80

replace \(x=-1\) and you will find C right away

Answer 81

on the left you will get \(-C\) and on the right you will get \(5(-1)^2+20\times -1+6=-9\)
which tells you \(C=9\)

Answer 82

now we can find B

Answer 83

wait... how do u know to use -1.. i wouldnt see that

Answer 84

because each term except \(Cx\) contains a factor of \(x+1\)

Answer 85

cause i'm looking at 5x^2 + 20x + 6 = .. all that stuff in the numerator

Answer 86

so if you replace \(x\) by -1 you will only be left with \(-C\) on the left