Question

integral sqrt(9-4x^2)/x dx

Answer 1

integral sqrt(9-4 x^2)/x dx = sqrt(9-4 x^2)-3 log(sqrt(9-4 x^2)+3)+3 log(x)+c (constant)

Answer 2

if you need the steps, you can ask for them right away in the question. If you merely need the answer, you could check out wolframalpha.com

Answer 3

u= 9-4x^2
du = -8x du
you don't have the -8 there, so divide both sides by -8,

you get
-1/8 du= x dx

Answer 4

Please, I need the steps

Answer 5

So your substitution is

u= 9-4x^2
-1/8 du = x dx

Answer 6

I am telling you the steps no?

Answer 7

I am not going to just do your work for you.

Answer 8

I am not allowed to. But I can help you... if you ar willing to work.

Answer 9

ok thanks

Answer 10

Do you think you understand why I am substituting u=9-4x^2
and -1/8 du = x dx ?

Answer 11

no

Answer 12

Have you learned U substitution?

Answer 13

Lets do this....
\[\large \int\limits_{ }^{ } \frac{\sqrt{9-4x^2}}{x}~dx\]\[u=9-4x^2\]\[\large \frac{du}{dx}=0-2\times4x^{2-1}=-8x\]
multiplying both sides times dx,
\[du = -8x~ dx\]

Answer 14

I think I did wrong exercise

Answer 15

lets do this, so that you can understand the concept.

Answer 16

or you don;t want to?

Answer 17

I'll proceed doing this, and let me know of any changes.

Answer 18

I see a trig sub.

Answer 19

ohh, it is 1/x there is no x inside.

Answer 20

I am blind again

Answer 21

okay

Answer 22

your u substitution gives a nice integrand for parital fractions

Answer 23

I prefer subing `9-4x^2 = u^2` though

Answer 24

lol yes!!

Answer 25

I haven't ever seen people do this, but don't see any restrictions against this. Awesome!
You again taught me something valuable.

Answer 26

then you would need to play around to get 1/x in term OF U.

Answer 27

you need an x in the numerator so you have to multiply x top and bottom

Answer 28

\[\large \int \frac{\sqrt{9-4x^2}}{x}~dx = \int \frac{\sqrt{9-4x^2}}{x^2}~xdx \]

Answer 29

I don't understand what did you really do, and don't think arquimus gets it

Answer 30

ohh, I see I thought of soemthing else... your square root didn't cover the entire thing, but that is the atex's fault.

Answer 31

what now it does? lagggs ... hatwe it. Sorry for interrupting you,

Answer 32

he multiplied x on top and bottom

Answer 33

yes, I got that part...

Answer 34

keep going with your substitution i think it goes smoothly, just chanege it to `9-4x^2 = u^2`

Answer 35

\[\large \int \frac{\sqrt{9-4x^2}}{x}~dx = \int \frac{\sqrt{9-4x^2}}{x^2}~xdx\]

\(9-4x^2 = u^2 \implies xdx = \dfrac{-u~du}{4}\)
\(x^2 = \frac{1}{4}(9-u^2)\)

the integral becomes

\[\large \int \dfrac{u^2}{u^2-9} ~du\]

Answer 36

\[3\ln \left| \frac{ 3-\sqrt{9-4x^2} }{ x } \right| + \sqrt{9-4x^2 } + C\]

Answer 37

I got this result, but I'm not okay

Answer 38

how did u get that ?

Answer 39

it looks correct except for a sign

Answer 40

by trigonometric substitution

Answer 41

http://www.wolframalpha.com/input/?i=%5Cint+%5Csqrt%289-4x%5E2%29%2Fx

Answer 42

if .... I think I made a mistake

Answer 43

I will continue my efforts .... bye