question_answer

Evaluate \[\int{{{e}^{2{{x}^{2\,\,+\,\,\text{In}\,\,x}}}}}dx.\]

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 \[2{{x}^{2}}=t\Rightarrow 4x\,dx=dt\]             \[\therefore \]      \[l=\frac{1}{4}\int{{{e}^{t}}\,\,dt=\frac{1}{4}}{{e}^{t}}+C\]                                    \[=\frac{1}{4}{{e}^{2{{x}^{2}}}}+C\]