If \[u=\int_{{}}^{{}}{{{e}^{ax}}\cos bx\ dx}\] and \[v=\int_{{}}^{{}}{{{e}^{ax}}\sin bx\ dx}\], then \[({{a}^{2}}+{{b}^{2}})({{u}^{2}}+{{v}^{2}})=\]

\[({{a}^{2}}-{{b}^{2}}){{e}^{2ax}}\]

\[({{a}^{2}}+{{b}^{2}}){{e}^{2ax}}\]

If \[u=\int_{{}}^{{}}{{{e}^{ax}}\cos bx\ dx}\] and \[v=\int_{{}}^{{}}{

1.	If \[u=\int_{{}}^{{}}{{{e}^{ax}}\cos bx\ dx}\] and \[v=\int_{{}}^{{}}{{{e}^{ax}}\sin bx\ dx}\], then \[({{a}^{2}}+{{b}^{2}})({{u}^{2}}+{{v}^{2}})=\]
A.	\[2{{e}^{ax}}\]
B.	\[({{a}^{2}}+{{b}^{2}}){{e}^{2ax}}\]
C.	\[{{e}^{2ax}}\]
D.	\[({{a}^{2}}-{{b}^{2}}){{e}^{2ax}}\]
Answer» D. \[({{a}^{2}}-{{b}^{2}}){{e}^{2ax}}\]

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If \[u=\int_{{}}^{{}}{{{e}^{ax}}\cos bx\ dx}\] and \[v=\int_{{}}^{{}}{{{e}^{ax}}\sin bx\ dx}\], then \[({{a}^{2}}+{{b}^{2}})({{u}^{2}}+{{v}^{2}})=\]

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