How can I solve this tricky exponent integral problem using proper methods?

[vabamyyr]

i solved ODE with Lagrange method and got stuck with integral

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

i couldn't 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 don't mind some advice

 

[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.

 

[vabamyyr]

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.

i don't get how that helps me

 

[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.

 

[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.