How Do You Integrate ae^(a^u) with Respect to u?

[Chaz706]

My brain is truly fried.

I know this:
$$\int e^u du = e^u +C$$

But what do I do if I get this:
$$\int ae^a^u du$$ ? Assuming a is a non-zero constant?

EDIT: Never mind! I figured this out backwards.

$$\frac {d}{du} e^a^u = ae^a^u du$$

Thus

$$\int ae^a^u du = e^a^u$$

 

[rune]

You can use substitution.
$$y=au\Rightarrow \frac{dy}{du}=a\Rightarrow du=\frac{dy}{a}$$
Therefore,
$$\int ae^{au}du=\int ae^{y}\frac{1}{a}dy=\int e^{y}dy=e^{y}+C=e^{au}+C$$

 

[thepatientmental]

+ C



It seems like you have already figured out the solution, but I will still provide a response for anyone else who may be struggling with this concept.

Firstly, it is important to remember that when solving an integral, we are essentially finding the antiderivative of the given function. In this case, the function is ae^a^u, where a is a non-zero constant.

To solve this integral, we can use the power rule for integration, which states that the integral of x^n is equal to (x^(n+1))/(n+1) + C. Applying this rule to our function, we get:

\int ae^a^u du = (ae^a^u)/(a+1) + C

However, this is not the final answer. We need to simplify the expression further by using the chain rule. The chain rule states that the derivative of a composite function f(g(x)) is equal to f'(g(x)) * g'(x). In our case, g(x) is a^u and f(u) is e^u. So, using the chain rule, we get:

\frac{d}{du} e^a^u = ae^a^u

Therefore, we can rewrite our integral as:

\int ae^a^u du = \frac{1}{a} * \frac{d}{du} e^a^u du = \frac{1}{a} * ae^a^u + C = e^a^u + C

And there we have it! Our final answer is e^a^u + C. I hope this helps clear up any confusion. Remember to always use the power rule and the chain rule when solving integrals. Keep practicing and it will become easier over time!