Question

i need help with integration of rational functions by partial fractions, the question is x+4/(x+1)(x+1)^2 please help

Answer 1

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{x+4}{(x+1)(x+1)^2}~dx}\) this?

Answer 2

for this one, you don't need partial fractions.

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{x+4}{(x+1)(x+1)^2}~dx}\)

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{x+4}{(x+1)^3}~dx}\) u=x+1, du=dx, x=u-1 ---> x+4=u-3

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{u-3}{u^3}~du}\)

break it up, and integrate using the power rule, then sub the x+1 for u back as you are done.

Answer 3

on top supposed to be u+3, not u-3. it is a typo, sorry.

Answer 4

with out the integral

Answer 5

\(\large\color{slate}{\displaystyle\frac{x+4}{(x+1)(x+1)^2}}\) you want partial fractions for this?

Answer 6

yeah the book has that

Answer 7

shoot its x+4/(x+1)^2 im sorry

Answer 8

its ok

\(\large\color{black}{ \displaystyle \frac{x+4}{(x+1)^2}= \frac{A}{x+1}+\frac{B}{(x+1)^2}}\)

Answer 9

this is the setup.

Answer 10

yeah that is what i have

Answer 11

solve for A and B, and re-write it

Answer 12

at first multiply every fractions time (x+1)^2

Answer 13

you will get
\(\large\color{black}{ \displaystyle x+4=A(x+1)+B}\)

Answer 14

so then the x value would be -1

Answer 15

yes, you can solve for B like that.

Answer 16

B=?

Answer 17

3

Answer 18

yes, B=3

Answer 19

now, you know that B=3, and lets choose an alternate x-value (say, x=1)

\(\large\color{black}{ \displaystyle x+4=A(x+1)+B}\)
\(\large\color{black}{ \displaystyle 1+4=A(1+1)+3}\)

Answer 20

solve for A.

Answer 21

or choosing x=0 is even better

Answer 22

oh ok there we go i wasnt putting 3 in

Answer 23

\(\large\color{black}{ \displaystyle 0+4=A(0+1)+3}\)

Answer 24

yes, you got to put that 3 for B, or else you have no way to find A:)

Answer 25

A=?

Answer 26

1

Answer 27

yes A=1

Answer 28

And let me verify it

Answer 29

here is the \(\scr\color{blue}{\href{ http://www.wolframalpha.com/input/?i=%28x%2B4%29%2F%28%28x%2B1%29%5E2%29+partial+fractions }{verification }}\).

Answer 30

oh, forgot another /

here is the \(\scr\color{blue}{\href{http:///www.wolframalpha.com/input/?i=%28x%2B4%29%2F%28%28x%2B1%29%5E2%29+partial+fractions}{verification }}\).

Answer 31

was supposed to have 3 times \ before wolfram, but just in case you now know how to make hyperlinks.

```
\(\scr\color{blue}{\href{http:///www.wolframalpha.com/input/?i=%28x%2B4%29%2F%28%28x%2B1%29%5E2%29+partial+fractions}{verification }}\).
```

Answer 32

bye

Answer 33

thanks it helped