[SOLVED] Integration by part problem with two different answers

**x3bnm** · November 24th 2009, 07:59 AM

I want to find the integration of following integral:

$\int x^3.e^{x^2}\, dx$

So i apply integration by part:

$u.v - \int v \,du\,\,$

$ Let \,\, u = x^3, du = 3x^2\,\, and \,\, dv = e^{x^2}, v = 2x.e^{x^2} $

So now th result is:

$\int x^3.e^{x^2}\, dx = 2x^4.e^{x^2} - \int\,\,6x^3.e^{x^2}\,\, dx$

$7\!\!\int x^3.e^{x^2}\, dx = 2x^4.e^{x^2}$

$\int x^3.e^{x^2}\, dx = \frac{2x^4.e^{x^2}}{7} + C$

Am i right? Because the answer to this problem is different back of the book.
Answer is $\frac{(x^2 -1).e^{x^2}}{2} + C$

The way i did it i can't find anything wrong with it. Can anyone kindly tell what
is wrong with my way of solving this problem?

**Drexel28** · November 24th 2009, 08:02 AM

Originally Posted by x3bnm

I want to find the integration of following integral:

$\int x^3.e^{x^2}\, dx$

So i apply integration by part:

$u.v - \int v \,du\,\,$

$ Let \,\, u = x^3, du = 3x^2\,\, and \,\, dv = e^{x^2}, v = 2x.e^{x^2} $

So now th result is:

$\int x^3.e^{x^2}\, dx = 2x^4.e^{x^2} - \int\,\,6x^3.e^{x^2}\,\, dx$

$7\!\!\int x^3.e^{x^2}\, dx = 2x^4.e^{x^2}$

$\int x^3.e^{x^2}\, dx = \frac{2x^4.e^{x^2}}{7} + C$

Am i right? Because the answer to this problem is different back of the book.
Answer is $\frac{(x^2 -1).e^{x^2}}{2} + C$

The way i did it i can't find anything wrong with it. Can anyone kindly tell what
is wrong with my way of solving this problem?

I think that we can clear some of the "clutter". Let $z=x^2$ so $dz=2x$ and this becomes $\frac{1}{2}\int z\cdot e^{z}dz$ . Which is easily done. Does that help?

EDIT: Ahh, did you differentiate when you should have integrated? For $v$

**x3bnm** · November 24th 2009, 08:13 AM

>EDIT: Ahh, did you differentiate when you should have integrated? For

Yes i did. That's the mistake i made. Thanks for finding it.

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