Mathematics

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

$e^{e+1}$

##### 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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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If $\displaystyle \int_{0}^{1}xe^{x^{2}}dx= \lambda \int_{0}^{1}e^{x^{2}}dx$ then $\lambda$
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${\int_{0}}^1 \sin^{-1}(\dfrac{2x}{1+x^2})dx=\dfrac{\pi}{2}-\log 2$