integration - Error in $intlimits_0^{infty}dx,frac {log^2 x}{a^2+x^2}$

This is my work for solving the improper integral$$I=\int\limits_0^{\infty}dx\,\frac {\log^2x}{x^2+a^2}$$I feel like I did everything write, but when I substitute values into $a$, it doesn’t match up with Wolfram Alpha.

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First substitute $x=\frac {a^2}u$ so$$\begin{align*}I & =-\int\limits_{\infty}^0du\,\frac {2\log^2a-\log^2 u}{x^2+a^2}\\ & =\int\limits_0^{\infty}du\,\frac {2\log^2a}{a^2+x^2}-\int\limits_0^{\infty}du\,\frac {\log^2 u}{a^2+x^2}\end{align*}$$Hence$$\begin{align*}I & =\int\limits_0^{\infty}du\,\frac {\log^2a}{a^2+x^2}\\ & =\frac {\log^2a}{a^2}\int\limits_0^{\pi/2}dt\,\frac {a\sec^2t}{1+\tan^2t}\\ & =\frac {\pi\log^2a}{2a}\end{align*}$$However, when $a=e$ Wolfram Alpha evaluates the integral numerically as$$I\approx 2.00369$$however the input that I arrived at evaluates numerically$$\frac {\pi}{2e}\approx0.5778$$Where did I go wrong? And how would you go about solving this integral?

Answer

Note
$$\log^2\left(\frac{a^2}{u}\right)\neq2\log^2a-\log^2u.$$