A calculus problem by Jonas Katona

By computing definite integrals here: $\int_{-\infty}^{\infty}\frac{x^2}{x^4+4}\,dx = \frac{1}{4}{\left(\frac{1}{2}\ln \mid (x-1)^2 +1 \mid - \ce{TAN}^{-1}(1-x))+\frac{1}{4}(\ce{TAN}^{-1}(x+1) -\frac{1}{2} \ln \mid (x+1)^2+1\mid)\right)}+\ce{C}$

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For computing boundaries see: $\begin{aligned} \int_{a}^{b} f(x) dx&= F(b) - F(a) \\&= \lim_{x\to b -}(F(x)) - \lim_{x \to a+}(F(x)) \end{aligned}$ .

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Therefore the answer of this is $\dfrac{\pi}{2}$ . Expressing in the from $\dfrac{\pi}{a}$ then to find the value of $a^a$ so let $a\ = 2$ . The value of $a^a$ is $4$ $\because a^a$ is $2^2\boxed{}$ .

Take note : you can convert the $\frac{\pi}{2}$ into degrees so the equivalent is $90^\circ$ .

$\LARGE \ce{FIN!!!!}$