Question

Write a value of $\int e^{2x^2+\ln x}dx$.

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \left(e^{2x^2+\ln x}\right)dx \\ =\int \left(e^{2x^2}\times e^{\ln x}\right)dx \\ =\int e^{2x^2}.xdx \\ \mathrm{Let}2x^2=t \\ \Rightarrow 4xdx=dt \\ \Rightarrow xdx=\frac{dt}{4} \\ \therefore I=\frac{1}{4}\int e^{t}dt \\ =\frac{1}{4}{e}^t+C \\ =\frac{1}{4}{e}^{2x^2}+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left(\mathit{\because }\mathit{t}\mathit{=}\mathit{2}{\mathit{x}}^{\mathit{2}}\right) \end{gathered}$$