How do you evaluate #\int e ^ { x } \sqrt { 9+ 2e ^ { x } } d x #?

2 Answers
Sep 12, 2017

#inte^xsqrt(9+2e^x)dx = (9+2e^x)^(3/2)/3#

Explanation:

#inte^xsqrt(9+2e^x)dx#

This isn't as bad as it looks, actually!

The first thing we can do is recognize that #sqrt(9+2e^x) = (9+2e^x)^(1/2)#

So our original equation turns into

#inte^x(9+2e^x)^(1/2)dx#

U-Substitution seems to be the best way to go.

Our #u# is the inside of the parenthesis.

#u=9+2e^x#

So,

#(du)/(dx)=2e^x#

We have #dx# in our original equation, don't we?
Let's solve for that, then.

#(du)/(dx)=2e^x# becomes #(du)/(2e^x)=dx#

Now that we have #u# and #dx# values, I am going to simplify the original equation.

#inte^x(u)^(1/2)(du)/(2e^x)#

You should immediately see the #e^x# on the top and the bottom. Let's cancel them!

#int(u)^(1/2)(du)/(2)#

We can take the #1/2# outside of the integral to get it out of the

#1/2int(u)^(1/2)(du)#

Now it looks much easier. (I hope)

The power rule tells us how to solve this.

Remember:

#f(x)=u^a#
#f'(x)=au^(a-1)#

Following that, we can solve our equation.

#1/2int(u)^(1/2)(du)#
#1/2(u)^(3/2)*(2/3)#

Simplifying this, our final answer is

#(u)^(3/2)/3#

Right?

Wrong!

Don't forget to substitute the real value for #u#!

#u=9+2e^x#

Our final answer is:

#(9+2e^x)^(3/2)/3#

You can check the answer by finding its derivative.

Sep 12, 2017

#inte^xsqrt(9+2e^x)dx=1/3(9+2x)^(3/2)+C#

Explanation:

we can do this by inspection

#inte^xsqrt(9+2e^x)dx#

#=int e^x(9+2x)^(1/2dx#

the outside is a constant multiplied by the derivative of the bracket, suggesting the integral is the bracket to the power +1

try #d/(dx)((9+2e^x)^(3/2))#

by the chain rule we have#=3/2xx2e^x(9+2e)^(1/2)#

#=3e^x(9+2e)^(1/2)#

comparing with the integral we have

#inte^xsqrt(9+2e^x)dx=1/3(9+2e^x)^(3/2)+C#