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Question: If \(\int_{}^{}{f(x)dx} = xe^{- \log|x|} + f(x),\) then \(f(x)\) is...

If ∫f(x)dx=xe−log⁡∣x∣+f(x),\int_{}^{}{f(x)dx} = xe^{- \log|x|} + f(x), then f(x)f(x) is

A

1

B

0

C

cexce^{x}

D

log⁡x\log x

Answer

cexce^{x}

Explanation

Solution

∫f(x)dx=xelog⁡∣1x∣+f(x)⇒∫f(x)dx=x∣x∣+f(x)\int_{}^{}{f(x)dx = xe^{\log\left| \frac{1}{x} \right|} + f(x) \Rightarrow \int_{}^{}{f(x)dx = \frac{x}{|x|} + f(x)}}

On differentiating both sides , we get

f(x)=0+f′(x)f(x) = 0 + f'(x)

We know ddx(ex)=ex,∴f(x)=cex\frac{d}{dx}(e^{x}) = e^{x},\therefore f(x) = ce^{x}.