\color{blue}{(1)}
\begin{align*} \int_{-\infty}^{+\infty}\frac{x^2+1}{x^4+1}{\rm d}x&=2\int_{0}^{+\infty}\frac{1+\frac{1}{x^2}}{\left(x-\frac{1}{x}\right)^2+2}{\rm d}x\\ &=2\int_{0}^{+\infty}\frac{{\rm d}\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^2+2}\\ &=2\int_{-\infty}^{+\infty}\frac{{\rm d}x}{x^2+2}\\ &=2\cdot\frac{1}{\sqrt{2}}\arctan\frac{x}{\sqrt{2}}\bigg|_{-\infty}^{+\infty}\\ &=\sqrt{2}\pi. \end{align*}
\color{blue}{(5)}
\begin{align*} \int_{-\infty}^{+\infty}\frac{x^2+1}{x^6+1}{\rm d}x&=2\left(\int_0^1\frac{x^2+1}{x^6+1}{\rm d}x+\int_{1}^{+\infty}\frac{x^2+1}{x^6+1}{\rm d}x\right)\\ &=2\left(\int_0^1\frac{x^2+1}{x^6+1}{\rm d}x+\int_{0}^{1}\frac{x^4+x^2}{x^6+1}{\rm d}x\right)\\ &=2\int_0^1 \frac{x^2+1}{x^4-x^2+1}{\rm d}x\\ &=2\int_0^1\frac{{\rm d}(x-\frac{1}{x})}{(x-\frac{1}{x})^2+1}\\ &=2\int_{-\infty}^{0}\frac{{\rm d}x}{x^2+1}\\ &=2\arctan x\bigg|_{-\infty}^0\\ &=\pi. \end{align*}
