Question

i am having problems with integration of rational functions by partial fractions. the question is integral x+4/(x^2+5x-6)

Answer 1

so the first step of the partial fraction decomposition you are looking for is factorization of the denominator. Do you know how to do so?

Answer 2

yes, what i have so far is \[X+4/(x+6)(x-1)\] which then you would have x+4=A(x-1)+B(x+6)

Answer 3

so as you can see you have a system of linear equations based on coefficient matching
A*x+B*x=x -> A+B=1
A*(-1)+B*(6)=4 -> -A+6B=4
correct?

Answer 4

ok i see what you did, but the answer i see in the back of the book is 1/7lnl(x+6)^2(x-1)^5l +C

Answer 5

do you want that method where you plug in values of x to solve for A and B?

Answer 6

\[\frac{x+4}{(x+6)(x-1)}=\frac{A}{x+6}+\frac{B}{x-1} \\ x+4=A(x-1)+B(x+6)\]
well we know we can make one term 0 on the right by plugging 1 for x

Answer 7

so try to solve for B using x=1

Answer 8

ok so B=5

Answer 9

let me check
1+4=A(1-1)+B(1+6)
5=A(0)+B(7)
5=0+7B
5=7B

Answer 10

B should be 5/7

Answer 11

ok i didnt think you would use A for it

Answer 12

we didn't use A since (1-1)=0
and A(0)=0

Answer 13

ok then, yeah i see

Answer 14

@freckles heaviside is awesome :o

Answer 15

ok so now you have
\[\frac{x+4}{(x+6)(x-1)}=\frac{A}{x+6}+\frac{B}{x-1} \\ x+4=A(x-1)+B(x+6) \\ \text{ where } B=\frac{5}{7}
\text{ So now we have } \\ x+4=A(x-1)+\frac{5}{7} (x+6)\]
now to make that one term on the right 0 replace x with -6 instead

Actually @inkyvovd I normally do the systems of equations way but I don't think op has learned that way

Answer 16

so would A=1

Answer 17

let's find out...
replacing x with -6...
\[x+4=A(x-1)+\frac{5}{7}(x+6) \\ -6+4=A(-6-1)+\frac{5}{7}(-6+6) \\ -6+4=A(-6-1)+\frac{5}{7}(0) \\ -2=A(-7)+0 \\ -2=A(-7)\]

Answer 18

so 2/7 ok, ok then, so the value you plug in is always the negative or positive of like (x+6) which goes to (-6+6)

Answer 19

i plugged in -6 becacuse that made x+6 zero
just like before when I plugged in 1, that would make x-1 zero

Answer 20

ust like on the previous problem
i plugged in 1 because it made two terms zero
then i plugged in 0 because it made one term zero

Answer 21

ok i see that makes the value searching for X or whatever variable you have easier

Answer 22

so after that you would do u substitution on those answers?

Answer 23

\[\frac{x+4}{(x+6)(x-1)}=\frac{A}{x+6}+\frac{B}{x-1}=\frac{\frac{2}{7}}{x+6}+\frac{\frac{5}{7}}{x-1}\]

Answer 24

then you would integrate that

Answer 25

1/7lnl(x+6)^2(x-1)^5l +C this is the answer

Answer 26

\[\frac{2}{7} \int\limits_{}^{} \frac{1}{x+6} dx+\frac{5}{7} \int\limits \frac{1}{x-1} dx\]

Answer 27

the factored out 1/7
and use power rule for log
and then product rule

Answer 28

you will find the following useful in showing their answer
ab+ac=a(b+c)
\[r \ln(x)=\ln(x^r) \text{ power rule } \\ \ln(a b)=\ln(a)+\ln(b) \text{ product rule }\]

Answer 29

so then the 2 in 2/7 would become the power for (x+6)
because of the power rule for ln

Answer 30

what do you have so far?

Answer 31

ln(x+6)^2(x-1)^5

Answer 32

\[\frac{2}{7} \int\limits\limits_{}^{} \frac{1}{x+6} dx+\frac{5}{7} \int\limits\limits \frac{1}{x-1} dx \\ \frac{2}{7} \ln|x+6|+\frac{5}{7}\ln|x-1| +C \\ \frac{1}{7}(2 \ln|x+6|+5 \ln|x-1|)+C \text{ by factoring } \\ \frac{1}{7}(\ln((x+6)^2)+\ln((x-1)^5))+C \text{ by power rule } \\ \frac{1}{7} (\ln((x+6)^2(x-1)^5)+C \text{ by product rule }\]

Answer 33

thank you again i do have one more question to post