Error done when computing $ I =\int_{0}^{+\infty} \frac{x^{2}+(1-\sqrt{2}) x+1}{x^{4}+1}\,\mathrm dx$ with the change of variable $x=\frac{1}{u}$

Question

Let us do the following change of variable: $x=\frac{1}{u}$ for $x>0$, then $\mathrm dx=-\frac{\mathrm du}{u^{2}}$ and \begin{align*} I &=\int_{0}^{+\infty} \frac{x^{2}+(1-\sqrt{2}) x+1}{x^{4}+1} \,\mathrm dx\\ &= \int_{0}^{+\infty} \frac{\frac{1}{u^{2}}+(1-\sqrt{2}) \frac{1}{u}+1}{\frac{1}{u^{4}}+1} \cdot\left(-\frac{\mathrm du}{u^{2}}\right) \\ &=-\int_{0}^{+\infty} \frac{u^{2}+(1-\sqrt{2}) u+1}{u^{4}+1} \,\mathrm du \\ &=-I. \end{align*} Thus $2I=0$, i.e. $I=0$. But the function inside the integral is positive, this integral can't equal zero. I computed the real value of I using another approach but i want to know why this one doesn't work, where is my error. I think i met all the conditions to do this change of variable:
$$t \mapsto u=\varphi(t)=\frac{1}{t}$$ is bijective over $(0;+\infty)$ and its derivative is continuous, moreover $$t \mapsto \frac{t^{2}+(1-\sqrt{2}) \cdot t+1}{t^{4}+1}$$ is defined over $\varphi(( 0 ;+\infty))=(0 ;+\infty)$.

Answer 1

Since you do $t=\frac1u$, the new integral should be$$\int_{+\infty}^0\frac{\frac1{u^2}+\left(1-\sqrt2\right)\frac1u+1}{\frac1{u^4}+1}\left(-\frac1{u^2}\right)\,\mathrm du=\int_0^{+\infty}\frac{\frac1{u^2}+\left(1-\sqrt2\right)\frac1u+1}{\frac1{u^4}+1}\left(\frac1{u^2}\right)\,\mathrm du.$$