# Thread: Difficulty associated with integration by part

1. ## Difficulty associated with integration by part

Hi,

I don't know what I'm doing wrong with computing this integral. The book's answer is $\int (x^3)(e^{x^2}) dx = \frac{1}{2}e^{x^2}\left(x^2 - 1) + c$

I'll show my steps and then elaborate on what I think I've done wrong so far towards the end of my post.

$\int (x^3)(e^{x^2}) dx$

$u = x^3 \Rightarrow \frac{du}{dx} = 3x^2$

$\frac{d}{dx} (e^{x^2}) \Rightarrow v = \frac{1}{2x} e^{x^2} + c$

$\int (x^3)(e^{x^2}) dx = (x^3)\left(\frac{1}{2x} e^{x^2}\right) - \int \left(\frac{1}{2x} e^{x^2}\right)(3x^2) dx$

$\int (x^3)(e^{x^2}) dx = \frac{1}{2}x^{2}e^{x^2} - \frac{3}{2} \int (x)(e^{x^2}) dx$

$\int (x^3)(e^{x^2}) dx = \frac{1}{2}x^{2}e^{x^2} - \frac{3}{2} \left[(x)\left(\frac{1}{2x}e^{x^2}\right) - \int e^{x^2} dx\right]$

$\int (x^3)(e^{x^2}) dx = \frac{1}{2}x^{2}e^{x^2} - \frac{3}{4}e^{x^2} - \frac{1}{2x}e^{x^2} + c$

$\int (x^3)(e^{x^2}) dx = \frac{1}{2}e^{x^2}\left(x^2 - \frac{1}{x} - \frac{3}{2}\right) + c$

Now, Wolfram Alpha confirms the book's answer is correct.

So I decided to treat $\int e^{x^2} dx$ as a seperate integral and I've got:

$\frac{d}{dx} (e^{x^2}) = 2xe^{x^2}$

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

However, Wolfram Alpha says:

$\int e^{x^2} dx = \frac{1}{2} \sqrt{\pi}\ erfi(x) + c$

And I don't know why?

2. ## Re: Difficulty associated with integration by part

Astar...

you are doing something totaly wrong . d(e^x2) is not (1/2x)(e^x^2)...there is your mistake..

d(e^x^2)=2xe^(x^2)

Anyway the Integral is as the book says and also wolfram is correct..Please revise your calculations

MINOAS

3. ## Re: Difficulty associated with integration by part

What are you on about? Are you differentiating or integrating?

$\frac{d}{dx}(e^{x^2}) = 2xe^{x^2} \Rightarrow \int e^{x^2}dx = \frac{e^{x^2}}{2x} + c$

That's the conclusion I came to so as to solve this seperate integral on its own, but, Wolfram Alpha says otherwise.

I've got signs mixed up as well as coefficients in my attempt, too.

4. ## Re: Difficulty associated with integration by part

One of the common mistakes people are doing in integrating by parts is thet they do not know how to manipulate the diferential dx..and find the two functions that take part in integrating by parts as invented by Brook Taylor.
[integral f(x)dg(x) = f(x)g(x)-integral g(x)df(x).

MULTIPLY THE INTEGRAL BY 2 AND DIVIDE IT AT THE SAME TIME by two GET ONe X FROM THE X^3 AND YOU GOT THE FUNCTION INSIDE THE DIFFERENTIAL ..IT IS SIMPLE...

Yess most C.A. S are programmed that way...
Anyway the integral is the one you found in the book anwers.

5. ## Re: Difficulty associated with integration by part

Forget it.

I'll post my solution when I've figured it out on my own.

6. ## Re: Difficulty associated with integration by part

$\frac{d}{dx}e^{x^{2}}=2xe^{x^{2}}$ so $\int 2xe^{x^{2}}dx=e^{x^{2}}+C.$

You can't now remove the $2x$ from the integral and shift it bottomside on the RHS of the equation, (which is what you appear to have done).

$\frac{1}{2}\int x^{2}.2xe^{x^{2}} dx$