# integration by parts

#### gomes

Am i doing something wrong? I cant seem to spot my mistake, the answer is in red (given by my teacher). #### skeeter

MHF Helper
$$\displaystyle \int x^3 \cdot e^{\frac{x^2}{2}} \, dx$$

$$\displaystyle t = \frac{x^2}{2}$$

$$\displaystyle dt = x \, dx$$

$$\displaystyle 2 \int x \cdot \frac{x^2}{2} \cdot e^{\frac{x^2}{2}} \, dx$$

substitute ...

$$\displaystyle 2 \int t \cdot e^{t} \, dt$$

now do parts

#### gomes

$$\displaystyle \int x^3 \cdot e^{\frac{x^2}{2}} \, dx$$

$$\displaystyle t = \frac{x^2}{2}$$

$$\displaystyle dt = x \, dx$$

$$\displaystyle 2 \int x \cdot \frac{x^2}{2} \cdot e^{\frac{x^2}{2}} \, dx$$

substitute ...

$$\displaystyle 2 \int t \cdot e^{t} \, dt$$

now do parts
thanks! just wondering, where did i go wrong in my first answer?

#### skeeter

MHF Helper
thanks! just wondering, where did i go wrong in my first answer?
$$\displaystyle e^{\frac{x^2}{2}}$$ does not have an elementary antiderivative ... so, it cannot be chosen as "dv"

• HallsofIvy

#### Random Variable

let $$\displaystyle u = x^{2}$$

and $$\displaystyle dv = x e^{x^{2}/2}$$

• HallsofIvy

#### HallsofIvy

MHF Helper
While, in differentiating something like $$\displaystyle e^{x^2}$$ you can just multiply by the derivative of $$\displaystyle x^2$$, to get $$\displaystyle 2xe^{x^2}$$, when integrating you cannot just divide by the derivative- it must already be in the integral.

• TheCoffeeMachine

#### gomes

thanks everyone, very helpful! erm, could someone clarify for me: how do i know if something has an elementary antiderivative or not?

MHF Helper
Last edited:

#### gomes

Thanks everyone! Similar Math Discussions Math Forum Date
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