How Do You Solve the Integral of \( e^{x^4} (x + x^3 + 2x^5) e^{x^2} \, dx \)?

[juantheron]

\(\displaystyle \displaystyle \int e^{x^4}\left(x+x^3+2x^5\right)\cdot e^{x^2}dx = \)

 

[Siron]

\(\displaystyle \displaystyle \int e^{x^4}\left(x+x^3+2x^5\right)\cdot e^{x^2}dx = \)

First, write the integral as
$$\int e^{x^4+x^2} (x+x^3+2x^5) dx = $$
Let $e^{x^4+x^2}=t \Rightarrow e^{x^4+x^2}(4x^3+2x)dx = dt$.

EDIT:
Thanks MarkFL, I didn't notice.

 

[MarkFL]

First, write the integral as
$$\int e^{x^4+x^2} (x+x^3+2x^5) dx = $$
Let $e^{x^4+x^2}=t \Rightarrow e^{x^4+x^2}(4x^3+2x)dx = dt$.

Here in the "Challenge Questions and Puzzles" forum, the OP has already solved the problem, and is posting it as a challenge for others to give a complete solution. :D

 

[anemone]

\(\displaystyle \displaystyle \int e^{x^4}\left(x+x^3+2x^5\right)\cdot e^{x^2}dx = \)

My solution:

Spoiler

Let $\displaystyle I=\int e^{x^2+x^4}\cdot(x+x^3+2x^5)dx$.

We see that

$\begin{align*}\displaystyle I&=\int e^{x^2+x^4}\cdot(x+x^3+2x^5)dx\\&=\dfrac{1}{x^2}\int x^2\cdot e^{x^2+x^4}\cdot(x+x^3+2x^5)dx\\&=\dfrac{1}{x^2}\int e^{\ln x^2+x^2+x^4}\cdot(x+x^3+2x^5)dx---(1)\end{align*}$

If we let $u=e^{\ln x^2+x^2+x^4}$, we get $\dfrac{x^2}{2}du=e^{\ln x^2+x^2+x^4}\cdot(x+x^3+2x^5) dx$.

Replacing this substitution to (1), the integral is then

$\begin{align*}\displaystyle I&=\dfrac{1}{x^2}\int e^{\ln x^2+x^2+x^4}\cdot(x+x^3+2x^5)dx\\&=\dfrac{1}{x^2}\int \dfrac{x^2}{2}du\\&=\int \dfrac{1}{2}du\\&=\frac{u}{2}\\&=\frac{e^{\ln x^2+x^2+x^4}}{2}\end{align*}$

$\therefore I=\dfrac{x^2(e^{x^2+x^4})}{2}$

 

[topsquark]

anemone: I think I spotted a flaw in your solution.

Spoiler

You wrote a statement of the form
\(\displaystyle \int f(x)~dx = \frac{1}{x^2} \int x^2~f(x)~dx\)

Under normal circumstances you can't multiply the outside of an integral by a function of x and multiply the inverse of it inside the integral. Or am I wrong in this specific case?

-Dan

 

[anemone]

anemone: I think I spotted a flaw in your solution.

Spoiler

You wrote a statement of the form
\(\displaystyle \int f(x)~dx = \frac{1}{x^2} \int x^2~f(x)~dx\)

Under normal circumstances you can't multiply the outside of an integral by a function of x and multiply the inverse of it inside the integral. Or am I wrong in this specific case?

-Dan

Hi Dan,

Spoiler

Hmm...but I was just, technically, multiplied a one $\dfrac{\cancel{x^2}}{\cancel{x^2}}=1$ to that integrand...I don't know if that's 100% permissible though. Thus, I stand corrected and I apologize if my solution is wrong.

 

[topsquark]

Spoiler

My thinking comes in two ways. First if \(\displaystyle \int f(x)~dx\) is a definite integral then \(\displaystyle \frac{1}{x^2} \int x^2 f(x)~dx\) would still depend on x, which would not be correct. As far as an indefinite integral is concerned let's do a simple example:
\(\displaystyle \int x^3~dx = \frac{x^4}{4} + C\)

\(\displaystyle \int x^3~dx = \frac{1}{x^2} \int x^2 \cdot x^3~dx = \frac{x^4}{6} + C\)

So they are not the same. However I don't know enough to say if there are cases where you can do this. I can say, though that the Taylor series of WA's answer about 0 and the Taylor series of your answer about 0 are not the same. So it looks like you can't do it in this case.

-Dan

 

[anemone]

Hi Dan again,

Spoiler

In your second example where \(\displaystyle \int x^3~dx = \frac{1}{x^2} \int x^2 \cdot x^3~dx = \frac{x^4}{6} + C\), you multiplied the antiderivative to the $x^2$ but in my solution, I took care of the multiplier before I started to integrate it on the second time. (Thinking)

 

[alyafey22]

Hi Dan again,

Spoiler

In your second example where \(\displaystyle \int x^3~dx = \frac{1}{x^2} \int x^2 \cdot x^3~dx = \frac{x^4}{6} + C\), you multiplied the antiderivative to the $x^2$ but in my solution, I took care of the multiplier before I started to integrate it on the second time. (Thinking)

A proof is a set of implications so each step has to be justified.

$$\frac{x^2}{x^2}=1$$

Does not imply that

$$\frac{1}{x^2}\int\,x^2\,dx = x+c $$