# Exponent integral problem

1. May 2, 2005

### vabamyyr

i solved ODE with Lagrange method and got stuck with integral

$$\int x^2 e^{\frac{1}{2}x^2+x} dx$$

i couldnt solve it with any method but combined and got the answer that it is

$$\int x^2 e^{\frac{1}{2}x^2+x} dx=C+(x-1)e^{\frac{1}{2}x^2+x}$$

the problem is that i want to do it with proper method, and show how it comes out. My mentor said that these types of integrals are "freaky" with little twist

Anyway, i dont mind some advice

Last edited: May 2, 2005
2. May 2, 2005

### dextercioby

U can do it using part integration.The derivative of the exponent is $x+1$ which u can obtain writing $x^{2}=x(x+1)-x$

Daniel.

3. May 3, 2005

### vabamyyr

i dont get how that helps me

4. May 3, 2005

### dextercioby

It does

$$\int x^{2}e^{\frac{1}{2}x^{2}+x} \ dx=\int \left(x^{2}+x-x\right) e^{\frac{1}{2}x^{2}+x} \ dx=\int x(x+1)e^{\frac{1}{2}x^{2}+x} \ dx-\int x e^{\frac{1}{2}x^{2}+x} \ dx$$
$$=xe^{\frac{1}{2}x^{2}+x}-\int e^{\frac{1}{2}x^{2}+x} \ dx-\int x e^{\frac{1}{2}x^{2}+x} \ dx$$
$$=x e^{\frac{1}{2}x^{2}+x}-\int (x+1)e^{\frac{1}{2}x^{2}+x} \ dx=x e^{\frac{1}{2}x^{2}+x}-e^{\frac{1}{2}x^{2}+x}+C$$

Daniel.

5. May 4, 2005

### vabamyyr

wow, that is very clever, at first i looked ur answer and couldn`t get one line but then i realized that the key moment was to crack int x(x+1)*e^...dx with simple method by taking u=x and dv=(x+1)*e^....dx and now i realize the beauty. So thank u very much for helping me on this one.

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