Integrate (xe^x)/(sqrt[1+e^x])

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In summary, the integration method for (xe^x)/(sqrt[1+e^x]) is substitution, specifically u-substitution. The value of the indefinite integral is (2sqrt[1+e^x]) + C. To solve a definite integral, the Fundamental Theorem of Calculus can be used. (xe^x)/(sqrt[1+e^x]) cannot be integrated by parts because the numerator is not a product of two functions. Other possible methods for integration include partial fractions and trigonometric substitutions.
  • #1
RoganSarine
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Homework Statement


Integrate: (xe^x)/(sqrt[1+e^x])


Homework Equations





The Attempt at a Solution



I tried substituting

u = e^x

So the equation became:

Integrate: ln|u|/sqrt[1+u]

But... That doesn't help me.



Trig substitution doesn't really help, neither does partial fractions. By parts doesn't work because of the fraction... So there's GOT to be a substitution I can't see.
 
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  • #2
Try [tex]u = \sqrt{1 + e^{x}}[/tex]
 

FAQ: Integrate (xe^x)/(sqrt[1+e^x])

What is the integration method for (xe^x)/(sqrt[1+e^x])?

The integration method for (xe^x)/(sqrt[1+e^x]) is substitution, specifically u-substitution. This involves substituting a new variable for the expression inside the parentheses, simplifying the integral, and then substituting back in the original variable.

What is the value of the indefinite integral of (xe^x)/(sqrt[1+e^x])?

The indefinite integral of (xe^x)/(sqrt[1+e^x]) is (2sqrt[1+e^x]) + C.

How do you solve a definite integral of (xe^x)/(sqrt[1+e^x])?

To solve a definite integral of (xe^x)/(sqrt[1+e^x]), you can use the Fundamental Theorem of Calculus. First, find the indefinite integral and then plug in the upper and lower limits of integration and subtract the two values.

Can (xe^x)/(sqrt[1+e^x]) be integrated by parts?

No, (xe^x)/(sqrt[1+e^x]) cannot be integrated by parts. This is because the numerator, xe^x, is not a product of two functions.

Are there any other methods for integrating (xe^x)/(sqrt[1+e^x])?

Yes, there are other methods for integrating (xe^x)/(sqrt[1+e^x]). Some possible methods include using partial fractions or trigonometric substitutions.

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