indefinite-integration

Question: $\int_{{}}^{{}}{{{e}^{2x+\log x}}}dx=$



1) $\frac{1}{4}(2x-1)+\frac{2}{x+1}+c$
2) $\frac{1}{4}(2x+1)+\frac{2}{x+1}+c$
3) $\frac{1}{2}(2x+1){{e}^{2x}}+c$
4) $\frac{1}{2}(2x+1){{e}^{2x}}+c$
Solution: Explanation: No Explanation
Integration by parts

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