Answer:
Let\[l=\int{{{e}^{2{{x}^{2}}\,\,+\ln \,\,x}}dx}\] \[=\int{{{e}^{2{{x}^{2}}}}\cdot {{e}^{\ln \,\,x}}dx}\] \[=\int{{{e}^{2{{x}^{2}}}}\cdot x\,\,dx}\] Put \[2{{x}^{2}}=t\Rightarrow 4x\,dx=dt\] \[\therefore \] \[l=\frac{1}{4}\int{{{e}^{t}}\,\,dt=\frac{1}{4}}{{e}^{t}}+C\] \[=\frac{1}{4}{{e}^{2{{x}^{2}}}}+C\]
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