if$$f'(x)=\frac{1}{1+x+x^2+x^3}$$,$$g'(x)=-\frac{x}{1+x+x^2+x^3}$$and $f(0)=g(0)$

Find value of $f(1)-g(1)$

Question

if$$f'(x)=\frac{1}{1+x+x^2+x^3}$$,$$g'(x)=-\frac{x}{1+x+x^2+x^3}$$and $f(0)=g(0)$

Find value of  $f(1)-g(1)$

anonymous · Answer

The Rational Zero Theorem suggests that (x-1) or (x+1) may be factors of the denominators.
Using polynomial long division (or synthetic division) indicates that (x+1) is a factor.
Use the resulting factored denominator with partial fraction decomposition to break both functions into something that can be integrated.
Integrate both functions with separately marked added constants (I used subscript f and g).
Set f(x)=g(x) substituting x=0 in for both functions to find what the constants should be.  You won't find the exact constants, but you will see how they relate to each other.
Now substitute x=1 into f(x)-g(x).

Would you like the thrill of discovery, or would you like to see my details?

helder_edwin · Answer

$$ f(1)-g(1)=\arctan 1$$

anonymous · Answer

Not exactly what I got... but close

anonymous · Answer

hey... I think you are right...arctan x with x=1... yup.
I incorrectly put x^2+1 into my argument, but it should be just x...nice!

anonymous · Answer

arctan 1=pi/4