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Question: Find \(\int {{e^{ - \log x}}dx} = \) \[ 1){\text{ }}{{\text{e}}^{ - \log x}} + c \\\ 2)\;x...

Find ∫e−log⁡xdx=\int {{e^{ - \log x}}dx} =

1) e−log⁡x+c 2)  xe−log⁡x+c 3) elog⁡x+c 4)  log∣x∣+c  1){\text{ }}{{\text{e}}^{ - \log x}} + c \\\ 2)\;x{{\text{e}}^{ - \log x}} + c \\\ 3){\text{ }}{{\text{e}}^{\log x}} + c \\\ 4)\;{\text{log}}\left| x \right| + c \\\
Explanation

Solution

Hint : The integration can be defined which denotes the summation of discrete data and it can be calculated to find the functions which will describe the area, displacement, volume, that occurs due to a collection of small data, which cannot be measured singularly. Here we will use different identities for the logarithmic and the power and exponents to simplify the given expression and then use the formula of integration to get the required value.

Complete step-by-step answer :
Take the given expression: ∫e−log⁡xdx\int {{e^{ - \log x}}dx}
By using the identity of the logarithm, alog⁡x=log⁡xaa\log x = \log {x^a} we write the above expression –
=∫elog⁡x−1dx= \int {{e^{\log {x^{ - 1}}}}dx}
Now, again by the property “e” exponential base and log cancel each other.
=∫x−1dx= \int {{x^{ - 1}}^{}dx}
By using the inverse function of the above expression, it can be written as –
=∫1xdx= \int {{{\dfrac{1}{x}}^{}}dx}
By using an identity –
=log⁡∣x∣+c= \log \left| x \right| + c
Hence, ∫e−log⁡xdx=log⁡∣x∣+c\int {{e^{ - \log x}}dx} = \log \left| x \right| + c
Therefore, from the given multiple choices – the fourth option is the correct answer.
So, the correct answer is “Option 4”.

Note : Don’t get confused between the identities for derivatives and the integration and apply accordingly. Remember the identity for the negative exponent and the inverse functions. Differentiation can be defined as the symbolization as the rate of change of the function, whereas integration suggests that the sum of the function over the range. Remember, Integration and derivation are inverses of each other.