integration by parts

**wolfhound** · February 18th 2010, 09:41 AM

Hello
How do I integrate x^4(1+x^2)^2 I know its: u dv/dx dx=uv- inte v dudx dx
so I know the integral of x^4 is (x^5)/5 but what the hell is the integral of dv/dx dx?
please help!

**e^(i*pi)** · February 18th 2010, 09:45 AM

Originally Posted by wolfhound

Hello
How do I integrate x^4(1+x^2)^2 I know its: u dv/dx dx=uv- inte v dudx dx
so I know the integral of x^4 is (x^5)/5 but what the hell is the integral of dv/dx dx?
please help!

I wouldn't do this via parts. As the exponent is low expand the binomial and then distribute the x^4

$(1+x^2)^2 = 1+2x^2+x^4$

$x^4(1+x^2)^2 = x^4(1+2x^2+x^4) = x^4 + 2x^6 + x^8$

$\int (x^4 + 2x^6 + x^8)\, dx = \int x^4\,dx + \int 2x^6\,dx + \int x^8\,dx$

Which is relatively simple to evaluate

================================================

If you want integration by parts anyway prepare for some hard work

$u = (1+x^2)^2 ........ du = 4x(1+x^2) = 4x+4x^3 \, dx$

$dv = x^4 .................. v = \frac{x^5}{5}$

$\frac{x^5(1+x^2)^2}{5} - \int \frac{x^5(4x+4x^3)}{5}$

and more integration by parts ensues

**wolfhound** · February 18th 2010, 09:53 AM

Thanks,
I am only new to integration and thought I was expected to use the product/part rule for integration
Thanks for your help

**wolfhound** · February 18th 2010, 10:20 AM

Can I see how to do it all by parts please?

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