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intsqrt(e^(x))dx=?...

`intsqrt(e^(x))dx=?`

A

`sqrt(e^(x))+C`

B

`2sqrt(e^(x))+C`

C

`(1)/(2)sqrt(e^(x))+C`

D

none of these

Text Solution

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The correct Answer is:
To solve the integral \(\int \sqrt{e^x} \, dx\), we can follow these steps: ### Step 1: Rewrite the Integral First, we rewrite the integral in a more manageable form: \[ \int \sqrt{e^x} \, dx = \int e^{x/2} \, dx \] ### Step 2: Integrate Now, we can integrate \(e^{x/2}\). The integral of \(e^{kx}\) is \(\frac{1}{k} e^{kx}\). Here, \(k = \frac{1}{2}\): \[ \int e^{x/2} \, dx = \frac{1}{\frac{1}{2}} e^{x/2} + C = 2 e^{x/2} + C \] ### Step 3: Rewrite the Result Finally, we can rewrite the result in terms of the square root: \[ 2 e^{x/2} + C = 2 \sqrt{e^x} + C \] ### Final Answer Thus, the integral \(\int \sqrt{e^x} \, dx\) is: \[ \int \sqrt{e^x} \, dx = 2 \sqrt{e^x} + C \] ---

To solve the integral \(\int \sqrt{e^x} \, dx\), we can follow these steps: ### Step 1: Rewrite the Integral First, we rewrite the integral in a more manageable form: \[ \int \sqrt{e^x} \, dx = \int e^{x/2} \, dx \] ...
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