Question

Is it possible to solve this integral by hand or by conventional methods?

`inte^(x^(2))*x^2dx`
I can use u sub, trig sub, integration by parts, tabular, or other conventional methods... I was told if this is not possible, just leave it be and use a calculator

Answer 1

You can't indeed.

Explanation:

You could have solved

`int xe^{x^2} dx`

since it is written in the form

`\int f'(x)g'(f(x))dx = g(f(x))`

up to some constants. But the one you wrote has no solutions in terms of standard functions.

Answer 2

This can be solved using integration by parts and u-substitution being aware of a common integral form that results in the so-called "error function". If you are not familiar with the error function (`"erf"(x)`) or its imaginary form (`"erf"i(x)`), then I would answer that this integral is not doable by conventional methods.

Answer 3

No, it is not.

Explanation:

Noting that:

`d/dx e^(x^2) = 2xe^(x^2)`

integrate by parts:

`int x^2 e^(x^2) dx = 1/2 int x d/dx(e^(x^2))dx`

`int x^2 e^(x^2) dx = (xe^(x^2))/2 - int e^(x^2)dx`

Now the gaussian integral :

`int e^(x^2)dx`

cannot be expressed in terms of elementary functions, so neither can the proposed function.

Answer 4

`int \ x^2e^(x^2) \ dx = sum_(n=0)^oo (x^(2n+3))/((2n+3)*n!)`

Explanation:

We seek:

`I = int \ x^2e^(x^2) \ dx`

There is no elementary solution in terms of traditional functions, however we can use:

`e^x = 1 + x + x^2/(2!) + x^3/(3!) + x^4/(4!) + ...`

so that:

`I = int \ x^2 \ {1 + (x^2) + (x^2)^2/(2!) + (x^2)^3/(3!) + (x^2)^4/(4!) + ... } \ dx`

`\ \ = int \ x^2 \ {1 + x^2 + x^4/(2!) + x^6/(3!) + x^8/(4!) + ... } \ dx`

`\ \ = int \ x^2 + x^4 + x^6/(2!) + x^8/(3!) + x^10/(4!) + ... \ dx`

`\ \ = x^3/3 + x^5/5 + x^7/(7*2!) + x^9/(9*3!) + x^11/(11*4!) + ... \ dx`

`\ \ = sum_(n=0)^oo (x^(2n+3))/((2n+3)*n!)`