Question

find an antiderivative of...

Answer 1

\[\frac{ x^2 }{ 4-x2 }\]

Answer 2

@terenzreignz is it not take int of the whole?

Answer 3

lol why ask me? :D

Answer 4

I'm still thinking. ;)

Answer 5

I think if we set it under the terminology of int to get the antiderivative, it is clearer to follow
and it is easier.

Answer 6

Might as well brandish the big guns...
@ParthKohli don't just stand there, do something :P

Answer 7

I'm not a big gun. Let me still try...

Answer 8

Well... that was interesting.

Answer 9

\[\dfrac{x^2}{4 - x^2} = \dfrac{x^2}{(2 + x)(2 - x)}\]\[= \dfrac{x}{2+x} \times \dfrac{x }{2 -x}\]I wonder if that helps.

Answer 10

How about we use longhand division first? If not, then a trig substitution might work. However I think a longhand division is imperative in this case of the integral, because even if we want to work with partial fraction decomposition later, we want to make sure that the grade of the denominator is higher than the grade of the numerator.

Answer 11

Got it!

Answer 12

x^2:(-x^2+4)=-1+(4/(-x^2+4)) if I didn't make any careless mistakes, now for the second term you can use pfd as shown by @ParthKohli above, split up the denominator into (2+x)(2-x)

Answer 13

I think I solved
\[\int\limits_{}^{}x^{2}/4-x^{2} \rightarrow x*x/4-x^{2} \rightarrow u=x , dV=x/4-x^{2}\]
\[V=-1/2\ln(4-x^{2}) , dx=1\] I think with twice use of part int we can salve this

Answer 14

edit: V=-1/2*ln(4-x^{2})

Answer 15

try \(x=tanh(u)\)

Answer 16

hyperbolic sub. might be easier.

Answer 17

trigonometric substitution
it just seems too good to be true
\[\large \int\frac{x^2}{4-x^2}dx\]
Let \(x=2\sin(\theta)\) \(dx = 2\cos(\theta)d\theta\)
\[\int \frac{(2\sin(\theta))^2}{4-(2\sin(\theta))^2}(2\cos(\theta)d\theta)\]\[\int\frac{4\sin^2(\theta)}{4(1-\sin^2(\theta))}(2\cos(\theta)d\theta)\]\[2\int \frac{\sin^2(\theta)}{\cos(\theta)}d\theta\]\[2\int\sec(\theta)\tan(\theta)d\theta\]\[2\sec(\theta)+C\]

Answer 18

I am sure hyperbolic sub might work, but I'd still recommend to use long hand division and partial fraction decomposition you will end up with:

\[\frac{ x^2 }{ -x^2+4 }=-1+\frac{ 1 }{ 2-x }+\frac{ 1 }{ 2+x }\]

Answer 19

easy integral isn't it?

Answer 20

wolf doesnt agree with your answer terrence :P

but long division by hand is tedious.

in my opinion; hack it with hyperbolic sub.

Answer 21

I must have made a mistake somewhere... then

Answer 22

but remember, that's not the final answer yet, that's still in terms of theta.

Answer 23

and again, emphasising - I must have made a mistake somewhere... if there is one

Answer 24

yeh it would require back substitution, seems elegant too to solve it that way.

Answer 25

@ksw19372 did you study hyperbolic functions already? If yes, it might be the easiest way to approach.

Answer 26

he/she should have; since its a calculus class.

http://en.wikipedia.org/wiki/Hyperbolic_function

Answer 27

not yet :(

Answer 28

i am stuck with the partial fraction part..

Answer 29

many people have hyperbolic functions later, I myself did cover them much later than the regular trig functions. In the end it's a matter of taste which way you want to go, I do believe that they all have their advantages/disadvantages.

For partial fraction decomposition you should end up with something like that:

4=A(2+x)+B(2-x)

So if you set x=2 you can read immediately that A=1
do the analog for x=-2

Answer 30

but where is the -1 from?

Answer 31

longhand division
x^2:(-x^2+4)=-1 + (4/(-x^2+4))

Answer 32

got the final ANSWER!
THANK YOU ALL!!