Difficulty associated with integration by part

**astartleddeer** · March 5th 2013, 08:41 AM

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?

**MINOANMAN** · March 5th 2013, 10:17 AM

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

**astartleddeer** · March 5th 2013, 11:02 AM

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.

**MINOANMAN** · March 5th 2013, 11:29 AM

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.

**astartleddeer** · March 5th 2013, 11:38 AM

Forget it.

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

**BobP** · March 5th 2013, 11:38 AM

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

Instead, write your integral as

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

and use the earlier result (twice) to integrate by parts.

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