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
$\displaystyle (1+x^2)^2 = 1+2x^2+x^4$
$\displaystyle x^4(1+x^2)^2 = x^4(1+2x^2+x^4) = x^4 + 2x^6 + x^8$
$\displaystyle \int (x^4 + 2x^6 + x^8)\, dx = \int x^4\,dx + \int 2x^6\,dx + \int x^8\,dx$
Which is relatively simple to evaluate
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If you want integration by parts anyway prepare for some hard work
$\displaystyle u = (1+x^2)^2 ........ du = 4x(1+x^2) = 4x+4x^3 \, dx$
$\displaystyle dv = x^4 .................. v = \frac{x^5}{5}$
$\displaystyle \frac{x^5(1+x^2)^2}{5} - \int \frac{x^5(4x+4x^3)}{5}$
and more integration by parts ensues