can you please please help with my integral calculus problem?

Question

if i try to integrate a fraction with a sum of two fourth powers in the denominator, there are two different ways: 1) using the partial fractions of the fraction or 2) using substitution. there is a difference in the answers (my answers are right, i checked them with friends, and if i plot the graphs leaving out the +c they are the same except for the y intercepts).

can you account for the difference in the answers?

thank you so much!

the question is to find the intergral of:
4x / (x^4+1)

the answer i got using partial fractions is:
2arctan[(root2)x - 1] - 2arctan[(root2)x+1] +c

and the one using substitution:
2arctanx^2 + c

Answer 1

Your answers differ by a constant. Remember that indefinite integration always has a +C with it to take into account the fact that the derivative of a constant is zero (and a function whose derivative is zero is a constant). This can easily happen when you do the same problem in two different ways. In fact, sometimes this is useful for finding identities: do an integral two different ways and find the constant difference between the answers.

In your case, there is an identity
arctan(A)-arctan(B)=
arctan[(A-B)/(1+A*B)].
If you let A=sqrt(2)x-1 and B=sqrt(2)x+1, you find that your first expression is
2arctan(-1/x^2)=
2arctan(x^2)-pi

Answer 2

I think both of your answers are correct--the constant is after all arbitrary for the indefinite integral. That +c can be altered to get the same y-intercept.

Though I'm too lazy to do this, you might try writing arctan in terms of natural logs (if you know Euler's fromula for trig functions in terms of exponentials, this might not be too bad). Then it might be easier to show that the two functions are the same but for a constant.

Answer 3

Did you do the back-substitution at the end?

It would help to see the equations and the resuts you got.

Answer 4

if you state the question in x terms i can help

did you mean double integral of 1/(x+y)^4

Answer 5

comprehensive squares (x^2-2x+2)= ((x^2-2x+a million) +a million) = ( x-a million)^2+a million INT 8 dx /( x-a million)^2+a million (x-a million) =Tan u dx= sec^2u du (x-a million)^2+a million = sec^2 u INT 8du = 8u = 8Arctan (x-a million) a million

Answer 6

Can you scan a paper with this problem and your solutions and tell me when you are ready to send it to me to my email address? I used to teach calculus, but without more information, I don't know how to help you