# Thread: Tricky intergal... or is it?!

1. ## Tricky intergal... or is it?!

I have not done calc in about 3 years and I'm currently in a calc class and I am stuck on a reveiw question:

integrate : $\displaystyle (x^4)e^(x^5)dx$

humm i cant seem to get it to display nice so but this is what im trying to write:

(x^4)e^(x^5)dx

I bet it's pretty simple, I forget alot :P Any help is greatly appriciated!

2. A simple u-substitution will do.

Let: $\displaystyle {\color{red}u = x^5} \ \Rightarrow \ du = 5x^4 \ dx \ \Leftrightarrow \ {\color{blue}\frac{du}{5} = x^4 \ dx }$

So your integral becomes: $\displaystyle \int {\color{blue}x^4}e^{\displaystyle {\color{red}x^5}} {\color{blue}dx} \ = \ \int e^{\displaystyle {\color{red}u}} \cdot {\color{blue}\frac{du}{5}}$

Factor out the $\displaystyle \tfrac{1}{5}$ and it should be a standard integral

3. Originally Posted by Krooger
I have not done calc in about 3 years and I'm currently in a calc class and I am stuck on a reveiw question:

integrate : $\displaystyle (x^4)e^(x^5)dx$

humm i cant seem to get it to display nice so but this is what im trying to write:

(x^4)e^(x^5)dx

I bet it's pretty simple, I forget alot :P Any help is greatly appriciated!
This can be solved by substitution, let $\displaystyle u=x^5$ and $\displaystyle du = 5x^4dx$ or $\displaystyle x^4dx = \frac{du}{5}$

then you can write the integral as $\displaystyle \frac{1}{5}\int{e^udu}$, integrate and substitute back in to get in terms of x again.

4. Awesome thank you for the help!