Question

evaluate integral 4ln(x+9)/x^2 dx

Answer 1

Why is it wrongg?

Answer 2

Lemme just make sure I took the same approach that you did, before I try to answer.

So you did this one by parts,
And then you end up with another integral that needs Partial Fraction Decomposition applied to it.

Yes? :D

Answer 3

ya, do u want me to upload myy work as well?

Answer 4

I think your partial fractions are incorrect, I'm getting the exact same answer just off by a constant factor on both of the fractions.

Answer 5

oh wait, i didnt do partial fractions, i just solved as using by parts

Answer 6

Oh hmm :D

Answer 7

Yah uploading your work might help then XD heh

Answer 8

i just let u= ln (x+9) , du = 1/x+9 , dv= x^-2 v = -x^-1

Answer 9

ok, but it is very messy, :/srry abt that

Answer 10

\[\huge =-\frac{1}{x}\ln(x+9)+\int\limits\limits \frac{1}{x(x+9)}dx\]

Answer 11

i dont think u can read that ..lol

Answer 12

lolll XD

Answer 13

i basically just calculated ln(x+9) / x^2
and then add the 4 back at the last step

Answer 14

How did you finish this part though?\[\huge \int\limits \frac{1}{x(x+9)}dx\]You have to break it down into 2 separate fractions :D Hmm I'm confused as to how you got past this step :3

Answer 15

1 break it into 1/x - 1/x+9..
oh..that's partial fraction..opps..

Answer 16

is it lol
and then integrate 1/x and 1/x+9

Answer 17

gives you lnx and ln(x+9)

Answer 18

it doesn't break up that way though :o
there should be a constant factor of 1/9 on each term i think.

That's what I came up with when i did it at least :D
You remember how to do good ole partial fractions? :)

Answer 19

not really... if there is a constant, it should be just on ln (x+9) but not ln x right

Answer 20

then basically it will be 4/9 ( the rest of the stuff)?

Answer 21

\[\large \frac{1}{x(x+9)}=\frac{A}{x}+\frac{B}{x+9}\]
\[\large 1=A(x+9)+Bx \qquad \text{Let x=0} \qquad 1=9A\]\[\large \text{Let x=-9} \qquad 1=-9B\]

Answer 22

You might wanna review partial fractions :D they're pretty helpful!
That's the only change i think you'll need to make, putting a 1/9 on both of the two last logs.

Answer 23

we are learning partial fractions right now @@ i get it when u write it out.. then it should be 4/ 81 then?

Answer 24

\[\frac{ 4 }{ 81 }\left( -\frac{ \ln(x+9) }{ x } \right)+\ln(x)-\ln(x+9)+C \]

Answer 25

the absolute sign doesnt matter right?

Answer 26

Hmm I'm not sure where the 81 is coming from :D\[\large \frac{A}{x}+\frac{B}{x+9} \qquad \rightarrow \qquad \frac{\frac{1}{9}}{x}+\frac{-\frac{1}{9}}{x+9}\]

Answer 27

u can rewrite that as 1/9 x- 1/9 (x+9) isnt it

Answer 28

and then take the constants out

Answer 29

Oh i see what you're doing,
\[\large =4\left(-\frac{1}{x}\ln(x+9)+\frac{1}{9}\ln |x| -\frac{1}{9}\ln |x+9|\right)+c\]
So if you pull the 1/9 out of each of the last 2 terms, you'll get this,\[\large =\frac{4}{9}\left(-\frac{9}{x}\ln(x+9)+\ln |x| -\ln |x+9|\right)+c\]

Answer 30

out of each term i mean* not just taken out of the last 2 terms, since it's all one big bracket.

Answer 31

oh right @@

Answer 32

I feel like i'm losing some basic math skills..:/

Answer 33

hah XD it happens!

Answer 34

but yea, thanks!! i have another ques (just need to check if i did it correctly), please take a look at that if u have time! gonna post it new