If int {x{e^{2x}},dx} is equal to {e^{2x}}f( | vedclass

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(C) To solve the integral $\int {x{e^{2x}}\,dx}$,we use the integration by parts formula: $\int {u\,dv} = uv - \int {v\,du}$.Let $u = x$ and $dv = {e^

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$(2x - 1)/4$

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(C) To solve the integral $\int {x{e^{2x}}\,dx}$,we use the integration by parts formula: $\int {u\,dv} = uv - \int {v\,du}$.Let $u = x$ and $dv = {e^{2x}}\,dx$.Then $du = dx$ and $v = \frac{{{e^{2x}}}}{2}$.Substituting these into the formula:$\int {x{e^{2x}}\,dx} = x \cdot \frac{{{e^{2x}}}}{2} - \int {\frac{{{e^{2x}}}}{2}\,dx}$$= \frac{{x{e^{2x}}}}{2} - \frac{1}{2} \cdot \frac{{{e^{2x}}}}{2} + C$$= \frac{{x{e^{2x}}}}{2} - \frac{{{e^{2x}}}}{4} + C$$= {e^{2x}}\left( {\frac{x}{2} - \frac{1}{4}} \right) + C$$= {e^{2x}}\left( {\frac{{2x - 1}}{4}} \right) + C$Comparing this with the given form ${e^{2x}}f(x) + C$,we get $f(x) = \frac{{2x - 1}}{4}$.

If $\int {x{e^{2x}}\,dx} $ is equal to ${e^{2x}}f(x) + C$,where $C$ is the constant of integration,then $f(x)$ is:

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