Need help, Integration of irrationals.

[sutupidmath]

Hi, i am trying so hard to find a way out of integrating this irrational function, but i can't jus figure out how to do it.
Here it is:
integ dx/(x^2 +x+1)(X^2 +x-1)^(1/2),
well here is what i tried. After some calculations i transformed the denominator to this form

integ dx/[(x+1/2)^2 +3/4][(x+1/2)^2 -5/4]^1/2, then i substituted x+1/2=t
where dx=dt, so i got

integ dt/(t^2 +3/4)(t^2 -5/4)^1/2 ,, then i tried a lot more but i can't just come to a solution, it keeps expanding instead of coming up to something. I think that from here i am not taking the right path.
So, can you guys give me some hints on how to go about integrating this??

thnx

P.S. The answer is in the textbook, i do not know whether it is correct or not, and it is:

I= 1/6^(1/2) ln[(3(x^2 +x-1))^(1/2)+(2x+1) 2^(1/2)]/[(3(x^2 +x-1))^(1/2)-(2x+1) 2^(1/2)]

 

[ZioX]

This is unreadable to a casual reader. If anyone helps you, be sure to thank them for taking their time to figure everything out.

If you want more help, please learn latex formating. :)

 

[HallsofIvy]

I think your integral is
$$\int \frac{dx}{(x^2+ x+ 1)\sqrt{x^2+ x- 1}}$$

A standard technique would be to complete the square in the radical and then look for a trig substitution.

 

[sutupidmath]

This is unreadable to a casual reader. If anyone helps you, be sure to thank them for taking their time to figure everything out.

If you want more help, please learn latex formating. :)

Yeah, i know. I really appologize for writing this way, but i don't know yet how to use latex. I guess i'll have to learn soon.

 

[sutupidmath]

I think your integral is
$$\int \frac{dx}{(x^2+ x+ 1)\sqrt{x^2+ x- 1}}$$

A standard technique would be to complete the square in the radical and then look for a trig substitution.

Yeah, i have completed the square in the radical, but i did not think of taking a trig supstitution. I'll try that tomorrow, couse it's too late now. Thnx for now, if i can't yet figure it out tomorrow, than surely i'll be back.

 

[sutupidmath]

Well, i tried a trig substituton, i took this substitution

t= [(5/4)^(1/2)] / sin u,

However the asnwer that i got is different from the one in the book, and also different from the one that "The integrator" ,an online version of mathematica as you may know, gives. But i think the idea works, i don't think i have made any mistakes in the process though, although it is quite likely.

 

[HallsofIvy]

$$x^2+ x- 1= x^2+ x+ \frac{1}{4}-\frac{1}{4}-1= x^2+ x+ \frac{1}{4}-\frac{5}{4}= (x+ \frac{1}{2})^2- \frac{5}{4}$$
A sin substitution would work for something of the form a²- usup]2[/sup] but this is of the form x²- usup]2[/sup]. (I just noticed that you have "/sin u" so that's really a csc substitution.)

Remembering that sin²x+ cos²x= 1, We can divide both sides by sin²x and get cot²x+ 1= csc²x or cot²x= csc²x- 1 which suggests the substitution $x+1= \sqrt{5}/2 csc(\theta)$. Then $dx= -\sqrt{5}/2 csc(\theta)cot(\theta) d\theta$. Of course, [itex]\sqrt{(x+1)^2- 5/4}= \sqrt{5}/2 cot(\theta). The hard part is the x²+ x+ 1 but fortunately the "x²+ x" is the same so we can complete the square in the same way: x²+x+1= (x+1)²+ 3/4. That polynomial becomes $(5/2)csc^2(\theta)$+ 3/4[/itex].

 

[sutupidmath]

$$x^2+ x- 1= x^2+ x+ \frac{1}{4}-\frac{1}{4}-1= x^2+ x+ \frac{1}{4}-\frac{5}{4}= (x+ \frac{1}{2})^2- \frac{5}{4}$$
A sin substitution would work for something of the form a²- usup]2[/sup] but this is of the form x²- usup]2[/sup]. (I just noticed that you have "/sin u" so that's really a csc substitution.)

Remembering that sin²x+ cos²x= 1, We can divide both sides by sin²x and get cot²x+ 1= csc²x or cot²x= csc²x- 1 which suggests the substitution $x+1= \sqrt{5}/2 csc(\theta)$. Then $dx= -\sqrt{5}/2 csc(\theta)cot(\theta) d\theta$. Of course, [itex]\sqrt{(x+1)^2- 5/4}= \sqrt{5}/2 cot(\theta). The hard part is the x²+ x+ 1 but fortunately the "x²+ x" is the same so we can complete the square in the same way: x²+x+1= (x+1)²+ 3/4. That polynomial becomes $(5/2)csc^2(\theta)$+ 3/4[/itex].

Well, this is almost exactly what i did, with the exception that i did not write the 1/sinu thing as cscu.

 

[VietDao29]

Well, i tried a trig substituton, i took this substitution

t= [(5/4)^(1/2)] / sin u,

However the asnwer that i got is different from the one in the book, and also different from the one that "The integrator" ,an online version of mathematica as you may know, gives. But i think the idea works, i don't think i have made any mistakes in the process though, although it is quite likely.

Ok, can you post your final answer here, so that we may verify it for you?

Btw, you should note that there are infinite number of anti-derivatives to f(x), all of which differ each other by a constant. E.g, -cos²(x), sin²x, and (-1/2)cos(2x) are all anti-derivatives of sin(2x), but they differ from each by a constant.

 

[sutupidmath]

Ok, can you post your final answer here, so that we may verify it for you?

Btw, you should note that there are infinite number of anti-derivatives to f(x), all of which differ each other by a constant. E.g, -cos²(x), sin²x, and (-1/2)cos(2x) are all anti-derivatives of sin(2x), but they differ from each by a constant.

Yeah, here it is, but i don't know latex formating, so it is going to be a little messy.

2/(6^1/2) log[2(2^1/2) -(3^1/2)arcsin(5^1/2)/(2x+1)]/[2(2^1/2) +(3^1/2)arcsin(5^1/2)/(2x+1)]
I don't really know whether this is right or not, it looks like it is but...
thnx in return

 

[Gib Z]

I'm not going to bother, but go to www.calc101.com and put that into the derivative finder. If its the original function, your home free. But you missed the whole point of VietDao29's post, you forgot the Constant :P

 

[sutupidmath]

I'm not going to bother, but go to www.calc101.com and put that into the derivative finder. If its the original function, your home free. But you missed the whole point of VietDao29's post, you forgot the Constant :P

NO, i got what VietDao29 meant about constant. I do understant that if F(x), and Q(x) are both primitive functions of f(x), than F(x)-Q(x)=C,
However in my case, i could not show the equivalence of the two solutions, the one that i got and the other which is on the textbook. Thnx for the link, i will try tha later, couse right now i am having some problems inputing it. But i got the idea about how to tackle these kinds of integrals. THNX.