Mathematics

If $$\int e^{x}(\imath nx+x\imath nx+1)dx=f(x)+cwhen f(1)=0,thenf(e)$$is equal to


ANSWER

$$e^{e+1}$$


SOLUTION
$$\int{{e}^{x} \left( \ln{x} + x \ln{x} + 1 \right) \; dx} = f{\left( x \right)} + c$$
Taking L.H.S.-
$$\int{{e}^{x} \left( \ln{x} + x \ln{x} + 1 \right) \; dx}$$
$$= \int{{e}^{x} \left( x \ln{x} + \left( 1 + \ln{x} \right) \right) \; dx}$$
$$= {e}^{x} \left( x \ln{x} \right) + c \quad \left[ \because \int{{e}^{x} \left( f{\left( x \right)} + f'{\left( x \right)} \right) = {e}^{x} f{\left( x \right)} + C \; dx} \right]$$
Comparing the above result with R.H.S., we get
$$f{\left( x \right)} = x {e}^{x} \ln{x}$$
$$\therefore f{\left( e \right)} = e . {e}^{e} \ln{e} = {e}^{e+1}$$
Hence $$f{\left( e \right)}$$ is equal to $${e}^{e+1}$$.
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Single Correct Medium Published on 17th 09, 2020
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