Integration by substitution: int x^4 e^(x^-2) dx
- Thread starter cmnalo
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As pka said, there is no elementary way.
I ran it through Maple and got:
\(\displaystyle \L\\-1/2\,{\frac {{x}^{3}{e}^{-{x}^{2}}}{\ln \left( e \right) }}-\frac{3}{4}\,{\frac {x{e}^{-{x}^{2}}}{ \left( \ln \left( e \right) \right) ^{2}}}+\frac{3}{8}\,{\frac {\sqrt {\pi }{\it erf} \left( \sqrt {\ln \left( e
\right) }x \right) }{ \left( \ln \left( e \right) \right) ^{\frac{5}{2}}}}\)
I ran it through Maple and got:
\(\displaystyle \L\\-1/2\,{\frac {{x}^{3}{e}^{-{x}^{2}}}{\ln \left( e \right) }}-\frac{3}{4}\,{\frac {x{e}^{-{x}^{2}}}{ \left( \ln \left( e \right) \right) ^{2}}}+\frac{3}{8}\,{\frac {\sqrt {\pi }{\it erf} \left( \sqrt {\ln \left( e
\right) }x \right) }{ \left( \ln \left( e \right) \right) ^{\frac{5}{2}}}}\)
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You might want to review the tutor's reply, in which it was noted that there is no elementary solution.cmnalo said:
Asking to be shown how to obtain an elementary antiderivative will not, I'm afraid, change the nature of this integral. It still can't be done. Sorry.
Eliz.
pka
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cmnola, this is a very frustrating problem. We have seen it many times before.
The functions \(\displaystyle \L xe^{ - x^2 } ,\quad x^3 e^{ - x^2 } ,\quad x^n e^{ - x^2 } \quad \left[ n \mbox{ odd} \right]\) all have simple anti-derivative. We get some these by using parts several times.
On the other hand, of these \(\displaystyle \L e^{ - x^2 } ,\quad x^2 e^{ - x^2 } ,\quad x^n e^{ - x^2 } \quad \left[n \mbox{ even} \right]\) none has a simple anti-derivative.
The functions \(\displaystyle \L xe^{ - x^2 } ,\quad x^3 e^{ - x^2 } ,\quad x^n e^{ - x^2 } \quad \left[ n \mbox{ odd} \right]\) all have simple anti-derivative. We get some these by using parts several times.
On the other hand, of these \(\displaystyle \L e^{ - x^2 } ,\quad x^2 e^{ - x^2 } ,\quad x^n e^{ - x^2 } \quad \left[n \mbox{ even} \right]\) none has a simple anti-derivative.
I apologise I must have mistyped the problem. The correct problem is:
∫x^4e^[(x^5)-2]dx
The answer is: (1/5)e^[(x^5)-2] + c
If I haven't wasn't too much of your time already it would still be great to get some help on how to set this problem up.
U= (x^5)-2
du= 5x^4 dx
∫x^4e^[(x^5)-2]dx
The answer is: (1/5)e^[(x^5)-2] + c
If I haven't wasn't too much of your time already it would still be great to get some help on how to set this problem up.
U= (x^5)-2
du= 5x^4 dx
Hello, cmnalo!
Do you know how to use Substitution?
You already have the set up . . . so what's stopping you?
Let \(\displaystyle u\:=\:x^5\,-\,2\;\;\Rightarrow\;\;du\:=\:5x^4dx\;\;\Rightarrow\;\;dx \:=\:\frac{du}{5x^4}\)
Substitute: \(\displaystyle \L\:\int x^4e^u\left(\frac{du}{5x^4}\right) \;= \;\frac{1}{5}\int e^udu \;=\;\frac{1}{5}e^u\,+\,C\)
Back-substitute: \(\displaystyle \L\:\frac{1}{5}e^{^{x^5-2}}\,+\,C\)
Do you know how to use Substitution?
You already have the set up . . . so what's stopping you?
Let \(\displaystyle u\:=\:x^5\,-\,2\;\;\Rightarrow\;\;du\:=\:5x^4dx\;\;\Rightarrow\;\;dx \:=\:\frac{du}{5x^4}\)
Substitute: \(\displaystyle \L\:\int x^4e^u\left(\frac{du}{5x^4}\right) \;= \;\frac{1}{5}\int e^udu \;=\;\frac{1}{5}e^u\,+\,C\)
Back-substitute: \(\displaystyle \L\:\frac{1}{5}e^{^{x^5-2}}\,+\,C\)