Question

The value of the integral
∫ 1 { x2 + x2013/(x2 + 2IxI +1)}dx
-1
(1) positive
(2) negative
(3) zero
(4) infinite




dx is

Answer 1

I suppose the given expression inside the integral is as below.

$I= \int\limits^{1}_{x=-1} [{x^2+\frac{x^{2013}}{x^2+2|x| +1}}] \, dx \\\\I= \int\limits^{1}_{-1} {x^2} \, dx + \int\limits^{1}_{-1} {x*\frac{(x^2)^{1006}}{x^2+2|x| +1}}} \, dx \\\\=\frac{1}{3}[x^3]_{-1}^{1}+\int\limits^{0}_{-1} {x*\frac{(x^2)^{1006}}{x^2+2|x| +1}}} \, dx +\int\limits^{1}_{0} {x*\frac{(x^2)^{1006}}{x^2+2|x| +1}}} \, dx \\\\I=\frac{2}{3}-\int\limits^{-1}_{0} {x*\frac{(x^2)^{1006}}{x^2+2|x| +1}}} \, dx +\int\limits^{1}_{0} {x*\frac{(x^2)^{1006}}{x^2+2|x| +1}}} \, dx \\\\I=2/3$

Let f(x) be function inside ONE Integral on the RHS.  Then let X= -x.

$F(x)=x*\frac{(x^2)^{1006}}{x^2+2|x| +1}\\\\I=2/3 - \int\limits^{1}_{X=0} {F(X)} \, dX + \int\limits^{1}_{x=0} {F(x)} \, dx\\\\Hence,\ I=\frac{2}{3}$

In the given integral $G(x) = \frac{x^{2013}}{x^2 + 2 |x| + 1}$ is anti symmetric function.  so its image wrt y axis.  So  G(x) = - G(-x).  Hence,  the area under the curve G(x) for x > 0 is equal and opposite to the area under the curve for  x < 0.