Yo lo que hice fue esto
\( \int_{1}^{e^a}{\frac{log(x)^{2}}{x^2}} \\
u=log(x)\\
du=\frac{1}{x}\\
\int_{}^{}{u^2.e^{-u}} \\
f(x)=u^2\ f'(x)=2u\\
g'(x)=e^{-u}\ g(x)=-e^{-u} \\
= u^2.-e^{-u} + \int_{}^{}{2u.e^{-u}} \\
\int_{}^{}{2u.e^{-u}}\\
f(x)=2u\ f'(x)=u\\
g'(x)=e^{-u}\ g(x)=-e^{-u} \\
\int_{}^{}{2u.e^{-u}}=2u.-e^{-u}+2 \int_{}^{}{e^{-u}}\\
\int_{1}^{e^a}{\frac{log(x)^{2}}{x^2}} \\= u^2.-e^{-u}+2u.-e^{-u}+2.-e^{-u}
\)
\( u^2.-e^{-u}+2u.-e^{-u}+2.-e^{-u} \\
-log(x)^2.-e^{-log(x)}+2log(x).-e^{-log(x)}+2.-e^{-log(x)} \)
Y cuando evaluas con 2 te termina dando la primera