Interesting Integral (1 Viewer)

[Riviet]

The other day, I was wondering how something like this could be integrated:
1/{(Ax+B)rt(Cx²+Dx+E)}

How would you go about doing it?

 

[haque]

this question has a lot of working and i can't put up my working because of the lack of symbols( i don't have that much time-even if i scanned it u guys won't be able to read my writing. this querstion also requires a lot of assumptions as no restrictions or conditions on the constant are specified-however i just dealt with the idea that the constants were greater than zero and thus took common factors out of square root signs without writing them as absolute values(which strictly speaking is the case: sq root x squared=abs value x.) my working probably had some careless error in it however i first took out a factor of C and completed the square in the square root bit, i assumed that m=E/C -D^2/4C^2 was greater than zero. i let u=x + D/2C and let K=B-AD/2C. after i worked through this i let u be sq root m tan@ from which it simplified down to 1/Asq rootC integral d@/sq root msin@ plus bcos @ where b=K/A. use the t=tan@/2 formula and resolve into partial fractions. after the integration is done, using the tan@ relationship from before form a quadratic in tan@/2. solving for this and ignoring the minus sign(assuming tan@/2 is positive as @/2 is acute) find tan@/2. after this u get an expression in "u". substitute u=x +D/2C or whatever and u get something like 1/Ab√mc/b<SUP>2</SUP>-c ln‌√m + √m+(x+D/2c)<SUP>2</SUP> –(x+D/2c)(√m/b +√m/b<SUP>2 </SUP>–1)/ √m + √m+(x+D/2c)<SUP>2</SUP>–(x+D/2c)(√m/b -√m/b<SUP>2 </SUP>–1)‌ + K<SUB>1<?xml:namespace prefix = o ns = "urn:schemas-microsoft-comfficeffice" /><o></o></SUB>

 

[KeypadSDM]

Integrals.wolfram? Or are you wondering how you would actually do it...

 

[Riviet]

It was posed by my teacher actually, he said that a 4 unit student should be able to integrate it and his two hints were to get rid of the linear part before completing the square later on.

 

[haque]

the working is a little shorter with that method but it isn't that much shorter.

 

[vafa]

this question has a lot of working and i can't put up my working because of the lack of symbols( i don't have that much time-even if i scanned it u guys won't be able to read my writing. this querstion also requires a lot of assumptions as no restrictions or conditions on the constant are specified-however i just dealt with the idea that the constants were greater than zero and thus took common factors out of square root signs without writing them as absolute values(which strictly speaking is the case: sq root x squared=abs value x.) my working probably had some careless error in it however i first took out a factor of C and completed the square in the square root bit, i assumed that m=E/C -D^2/4C^2 was greater than zero. i let u=x + D/2C and let K=B-AD/2C. after i worked through this i let u be sq root m tan@ from which it simplified down to 1/Asq rootC integral d@/sq root msin@ plus bcos @ where b=K/A. use the t=tan@/2 formula and resolve into partial fractions. after the integration is done, using the tan@ relationship from before form a quadratic in tan@/2. solving for this and ignoring the minus sign(assuming tan@/2 is positive as @/2 is acute) find tan@/2. after this u get an expression in "u". substitute u=x +D/2C or whatever and u get something like 1/Ab√mc/b<SUP>2</SUP>-c ln‌√m + √m+(x+D/2c)<SUP>2</SUP> –(x+D/2c)(√m/b +√m/b<SUP>2 </SUP>–1)/ √m + √m+(x+D/2c)<SUP>2</SUP>–(x+D/2c)(√m/b -√m/b<SUP>2 </SUP>–1)‌ + K<SUB>1fficeffice" /><O></O></SUB>

This is absolutely wrong...!

 

[joesmith1975]

wolfram gave me this

pretty messed up indeed

 

[AppleXY]

intergals at wolfram is the ultimate

 

[Slidey]

The other day, I was wondering how something like this could be integrated:
1/{(Ax+B)rt(Cx²+Dx+E)}

If you can split it into partial fractions as u/(Ax+B) + w/(sqrt(Cx^2+Dx+E), then you could easily solve the integral using two standard integrals. Although those constants everywhere would be EXTREMELY messy.

How can you split it up into partial fractions?

joesmith: Integrator doesn't recognise rt as the root function. It treated it like a constant.

 

[joesmith1975]

sorry entered wrong

Question: how do u make the picture into the post so dont have to open it ??

 

[haque]

justify your statement that this is absolutely wrong vafa as my answer is in the form of the wolfram answe-unless you are saying wolfram is wrong?

 

[haque]

as i have alread6y mentioned before i may have made a careless error as i had done it quickly and roughly-however the basic format of the answer is correct-there is also nothing wrong with the methods i employed although complex-maybe u didn't understand what i wrote the first time.