Whats the integral of dx/(1+x^4) ?

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Whats the integral of dx/(1+x^4) ?

Posted 263 day ago

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Answers (6)

Read your answer from tables of integrals involving (x^4 + a^4)

Let s=square root of 2.

Integral dx/(x^4 + 1) =
(1/4s)ln(x^2+ xs + 1)/(x^2-xs+ 1) -
(1/2s)[tan^-1(1-xs)-tan^-1(1+xs)]

Posted 217 day ago

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asad.taj

[image width=425]file:///C:/Documents%20and%20Settings/Asad/My%20Documents/graph.jpg[/image]

Posted 231 day ago

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Amar Soni

The solution of this is very lengthy it cannot be typed here.Send your e-mail I will send you a solution.

Posted 250 day ago

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atta

The integral of dx / (1 + x^4) does not seem to fit any of the integrands for the differential and integral formulas. That which can be done to simplify the integral is to reduce 1 + x^4 into a factor of irreducible primes such as (x^2 + 1)(x - 1)(x + 1) = (x^2 + 1)(x^2 - 1) => (x^2 + 1)^2 - (sqrt(2)x)^2. Thus, this integral can be re-written as f(x) = dx / (1 + x^4) = dx / (x^2 + 1)^2 - (sqrt(2)x)^2 => S dx / u^2 - a^2 = 1/2a ln|u - a / u + a| + C. Thus, by integrating the rational function yields an integral of f'(x) = S 1du / u^2 - a^2 = 1/2(sqrt(2)x) ln|(x^2 - sqrt(2)x + 1) / (x^2 + sqrt(2)x + 1)| + C

J.C

Posted 262 day ago

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Doctor Bob

http://www.ma.utexas.edu/users/kawasaki/mathPages.dir/ may be of some help to you. Scroll down to INTEGRAL.

Posted 262 day ago

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steiner1745

This is very messy to do in full. But let me get you started.
Note that x^4 + 1 = (x^2 + √2 x + 1)(x^2 -√2x + 1).
Now break 1/[ (x^2 + √2 x + 1)(x^2 -√2x + 1)] into
partial fractions and integrate. You will get 2
terms with logarithms and 2 with arctangents.
Good luck!

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