How do you integrate #int e^xsqrt(1-e^x)dx#?

Question

Luke Alvarez · Accepted Answer

#(-2(1-x^e)^(3/2))/3+C#

There is one thing to remember when you first look at an integral problem involving substitution: Which section looks like the derivative of the other. Since we are dealing with exponental function, this is the simplest problem there is. The derivative of #e^x=e^x#.

Given the existence of a square root, any addition to the root will necessitate the application of a chain rule, making this problem challenging. Using the substitution, we will make:

#u=1-e^x# #du=-e^x dx#

Consequently, following the substitution, the final integral is

#-intsqrt(u) du# #-intu^(1/2)du#

After applying the integral, multiply the resultant number by the exponent, and divide it.

#(-2u^(3/2))/3+C#

Once you replace the u from the initial scenario, you have your definitive response.

#(-2(1-x^e)^(3/2))/3+C#

Adam Adkins · Answer

undefined To integrate $ \int e^{x\sqrt{1-e^x}} \, dx $, you can use the substitution method. Let $ u = \sqrt{1 - e^x} $, then $ u^2 = 1 - e^x $ and $ e^x = 1 - u^2 $. Taking the derivative with respect to $ x $ gives $ du/dx = -\frac{1}{2\sqrt{1 - e^x}}e^x $, which simplifies to $ du = -\frac{1}{2u}e^x \, dx $.

Substituting these expressions into the integral, we get:
$$ \int e^{x\sqrt{1-e^x}} \, dx = -2\int e^{-x} \, du $$

This is now a simpler integral to evaluate. Integrating $ e^{-x} $ with respect to $ u $ gives $ -e^{-x} $, so:
$$ -2\int e^{-x} \, du = -2(-e^{-x}) + C = 2e^{-x} + C $$

Thus, the integral of $ \int e^{x\sqrt{1-e^x}} \, dx $ equals $ 2e^{-x} + C $, where $ C $ is the constant of integration.