Does $f(x)=1/(1+x^{100})$ have an antiderivative on $(-1,1)$? Its clear that $f$ is Riemann integrable, so if an antiderivative were to exist, it would have to agree to this up to a constant. How would I proceed?
 A: Since $f$ is continuous in $]-1,1[$, it has an anti-derivative there, defined by $$F(x) = \int_{a}^x \frac{1}{1+t^{100}}\,{\rm d}t,$$where $a \in ]-1,1[$.
Recall the Fundamental Theorem of Calculus.
A: I am puzzled that you would ask.  It is clearly continuous on (-1, 1) so certainly integrable there. 
(If the question had been "on (-1, 1]" it would be harder.)
A: This can be written as
$$-\dfrac{\displaystyle\sum_{\left\{{\omega}:\>{\omega}^{16}-{\omega}^{12}+{\omega}^8-{\omega}^4+1=0\right\}}\frac{\left({\omega}^{12}-2{\omega}^8+3{\omega}^4-4\right)\ln\left(\left|x-{\omega}\right|\right)}{16{\omega}^{15}-12{\omega}^{11}+8{\omega}^7-4{\omega}^3}}{25}-\dfrac{\displaystyle\sum_{\left\{{\omega}:\>{\omega}^{80}-{\omega}^{60}+{\omega}^{40}-{\omega}^{20}+1=0\right\}}\frac{\left({\omega}^{60}-2{\omega}^{40}+3{\omega}^{20}-4\right)\ln\left(\left|x-{\omega}\right|\right)}{80{\omega}^{79}-60{\omega}^{59}+40{\omega}^{39}-20{\omega}^{19}}}{5}+\dfrac{\frac{\ln\left(\left|x^2+\sqrt{2}x+1\right|\right)}{2^\frac{5}{2}}-\frac{\ln\left(\left|x^2-\sqrt{2}x+1\right|\right)}{2^\frac{5}{2}}+\frac{\arctan\left(\frac{2x+\sqrt{2}}{\sqrt{2}}\right)}{2^\frac{3}{2}}+\frac{\arctan\left(\frac{2x-\sqrt{2}}{\sqrt{2}}\right)}{2^\frac{3}{2}}}{25}+C$$
Which simplifies to:
$$-\dfrac{2^\frac{5}{2}\left(\displaystyle\sum_{\left\{{\omega}:\>{\omega}^{16}-{\omega}^{12}+{\omega}^8-{\omega}^4+1=0\right\}}\frac{\left({\omega}^{12}-2{\omega}^8+3{\omega}^4-4\right)\ln\left(\left|x-{\omega}\right|\right)}{16{\omega}^{15}-12{\omega}^{11}+8{\omega}^7-4{\omega}^3}\right)+5{\cdot}2^\frac{5}{2}\left(\displaystyle\sum_{\left\{{\omega}:\>{\omega}^{80}-{\omega}^{60}+{\omega}^{40}-{\omega}^{20}+1=0\right\}}\frac{\left({\omega}^{60}-2{\omega}^{40}+3{\omega}^{20}-4\right)\ln\left(\left|x-{\omega}\right|\right)}{80{\omega}^{79}-60{\omega}^{59}+40{\omega}^{39}-20{\omega}^{19}}\right)-\ln\left(\left|x^2+\sqrt{2}x+1\right|\right)+\ln\left(\left|x^2-\sqrt{2}x+1\right|\right)-2\arctan\left(\frac{2x+\sqrt{2}}{\sqrt{2}}\right)-2\arctan\left(\frac{2x-\sqrt{2}}{\sqrt{2}}\right)}{25{\cdot}2^\frac{5}{2}}+C$$
Between -1 and 1, this is approximately $1.986299765458299$
Courtesy of http://www.integral-calculator.com
