An odd integral

Section I. Finding Global Maxima

Finding global maxima of $f(x)$ , $\begin{array}{rl} f'(x) &= \dfrac{\left(x^4 + 4\right) \cdot 2x - x^2 \cdot 4x^3}{\left(x^4 + 4\right)^2}\\ &= \dfrac{-2x^5 + 8x}{\left(x^4 + 4\right)^2}\\ 0 &= -2x\left(x^4 - 4\right)\\ \Longrightarrow \qquad x&= \pm\sqrt{2}\\ \end{array}$ which show that the global maxima are $\left(\sqrt{2}, \dfrac{1}{4}\right)$ and $\left(-\sqrt{2}, \dfrac{1}{4}\right)$ .

Section II. Integration

Note: Instead of doubling the halved region, the best way to obtain the close form is to compute the integral the long way. Doubling the halved region gives the same results, but different values. You need to then match the corresponding terms with your answer to find the matching values.

Call $t(x) = \dfrac{1}{4}$ the equation of the line tangent to two points of $f(x)$ . Since $t(x) \geq f(x)$ at $-\sqrt{2} \leq x \leq \sqrt{2}$ , the area integral is $\int_{-\sqrt{2}}^{\sqrt{2}} \left(\dfrac{1}{4} - \dfrac{x^2}{x^4 + 4}\right)\,dx$ which is also $\int_{-\sqrt{2}}^{\sqrt{2}} \dfrac{1}{4}\,dx - \int_{-\sqrt{2}}^{\sqrt{2}} \dfrac{x^2}{x^4 + 4}\,dx$ Clearly, $\int_{-\sqrt{2}}^{\sqrt{2}} \dfrac{1}{4}\,dx = \dfrac{\sqrt{2}}{2} = \dfrac{1}{\sqrt{2}}$ which yields $\dfrac{1}{\sqrt{2}} - \int_{-\sqrt{2}}^{\sqrt{2}} \dfrac{x^2}{x^4 + 4}\,dx$ For the remaining integral, since $x^4 + 4 = \left(x^2 + 2x + 2\right)\left(x^2 - 2x + 2\right)$ by partial decomposition, we have $-\int_{-\sqrt{2}}^{\sqrt{2}} \left(\dfrac{x}{4\left(x^2 - 2x + 2\right)} - \dfrac{x}{4\left(x^2 + 2x + 2\right)}\right)\,dx = \dfrac{1}{4} \int_{-\sqrt{2}}^{\sqrt{2}} \left(\dfrac{x}{x^2 + 2x + 2} - \dfrac{x}{x^2 - 2x + 2}\right)\,dx$ For each decomposed fraction, express $x = x + (1 - 1)$ and $\begin{array}{rlcrl} x^2 + 2x + 2 &= \left(x^2 + 2x + 1\right) + 1 & & x^2 - 2x + 2 &= \left(x^2 - 2x + 1\right) + 1\\ &= \left(x + 1\right)^2 + 1 & & &= \left(x - 1\right)^2 + 1 \end{array}$ so $\dfrac{1}{4} \int_{-\sqrt{2}}^{\sqrt{2}} \left(\dfrac{x + 1}{x^2 + 2x + 2} - \dfrac{1}{\left(x + 1\right)^2 + 1} - \left(\dfrac{x - 1}{x^2 - 2x + 2} + \dfrac{1}{\left(x - 1\right)^2 + 1}\right)\right) \,dx = \dfrac{1}{4} \int_{-\sqrt{2}}^{\sqrt{2}} \left(\dfrac{x + 1}{x^2 + 2x + 2} - \dfrac{x - 1}{x^2 - 2x + 2} - \left(\dfrac{1}{\left(x + 1\right)^2 + 1} + \dfrac{1}{\left(x - 1\right)^2 + 1}\right)\right)\,dx$ Integrating by substitution, $\left.\dfrac{1}{4}\left(\dfrac{1}{2}\left(\ln\left(x^2 + 2x + 2\right) - \ln\left(x^2 - 2x + 2\right)\right) - \left(\arctan\left(x + 1\right) + \arctan\left(x - 1\right)\right)\right)\right|_{x = -\sqrt{2}}^{x = \sqrt{2}}$ which by $\arctan(p) + \arctan(q) = \arctan\left(\dfrac{p + q}{1 - pq}\right)$ gives $\begin{array}{rl} & \left.\dfrac{1}{4}\left(\dfrac{1}{2}\ln\left(\dfrac{x^2 + 2x + 2}{x^2 - 2x + 2}\right) - \arctan\left(\dfrac{2x}{2 - x^2}\right)\right)\right|_{x = -\sqrt{2}}^{x = \sqrt{2}}\\ \Longrightarrow & \left.\dfrac{1}{4}\left(\dfrac{1}{2}\ln\left(\dfrac{x^2 + 2x + 2}{x^2 - 2x + 2}\right) - \arcsin\left(\dfrac{2x}{\sqrt{x^4 + 4}}\right)\right)\right|_{x = -\sqrt{2}}^{x = \sqrt{2}} \end{array}$ So by Fundamental Theorem of Calculus, $\begin{array}{rl} & \dfrac{1}{4}\left(\dfrac{1}{2}\left(\ln\left(\dfrac{\sqrt{2}^2 + 2\sqrt{2} + 2}{\sqrt{2}^2 - 2\sqrt{2} + 2}\right) - \ln\left(\dfrac{\sqrt{2}^2 - 2\sqrt{2} + 2}{\sqrt{2}^2 + 2\sqrt{2} + 2}\right)\right) - \left(\arcsin\left(\dfrac{2\sqrt{2}}{\sqrt{\sqrt{2}^4 + 4}}\right) - \arcsin\left(-\dfrac{2\sqrt{2}}{\sqrt{\left(-\sqrt{2}\right)^4 + 4}}\right)\right)\right)\\ =& \dfrac{1}{4}\left(\dfrac{1}{2}\left(\ln\left(\dfrac{2 + \sqrt{2}}{2 - \sqrt{2}}\right) - \ln\left(\dfrac{2 - \sqrt{2}}{2 + \sqrt{2}}\right)\right) - \left(\arcsin(1) - \arcsin(-1)\right)\right)\\ =& \dfrac{1}{4}\left(\dfrac{1}{2}\ln\left(\dfrac{\left(2 + \sqrt{2}\right)^2}{\left(2 - \sqrt{2}\right)^2}\right) - \pi\right)\\ =& \dfrac{1}{4}\left(\dfrac{1}{2}\ln\left(\dfrac{6 + 4\sqrt{3}}{6 - 4\sqrt{3}}\right) - \pi\right)\\ \end{array}$ Since $\tanh^{-1}(z) = \dfrac{1}{2}\ln\left(\dfrac{1 + z}{1 - z}\right)$ Inside $\ln$ , divide each term by $6$ , so $\dfrac{1}{2}\ln\left(\dfrac{1 + {\color{#3D99F6}\left(\dfrac{2\sqrt{2}}{3}\right)}}{1 - {\color{#3D99F6}\left(\dfrac{2\sqrt{2}}{3}\right)}}\right) = \tanh^{-1}{\color{#3D99F6}\left(\dfrac{2\sqrt{2}}{3}\right)}$ So, $\begin{array}{rl} \dfrac{1}{4}\left(\tanh^{-1}\left(\dfrac{2\sqrt{2}}{3}\right) - \pi\right) &= \dfrac{1}{4}\tanh^{-1}\left(\dfrac{2\sqrt{2}}{3}\right) - \dfrac{\pi}{4}\\ &= \dfrac{1}{4}\tanh^{-1}\left(\dfrac{2\sqrt{2}}{3}\right) - \arcsin\left(\dfrac{1}{\sqrt{2}}\right) \end{array}$ Thus, the area of the integral is $\boxed{\dfrac{1}{4}\tanh^{-1}\left(\dfrac{2\sqrt{2}}{3}\right) - \arcsin\left(\dfrac{1}{\sqrt{2}}\right) + \dfrac{1}{\sqrt{2}}}$ where $a = 4, d = 3, b = c = e = f = 2\Longrightarrow a + b + c + d + e + f = \boxed{15}$