# Integration

• Nov 13th 2012, 02:17 PM
NettieL
Integration
Hi there,

just wondering if someone could help with this: integrate x^3(1+x^8)^-1(dx)

if u=(x^4)^2

http://www3.wolframalpha.com/Calcula...=46&w=253&h=40how does the x^3 become 1 in this equation?
• Nov 13th 2012, 04:04 PM
skeeter
Re: Integration
$\int \frac{x^3}{1+(x^4)^2} \, dx$

$u = x^4$

$du = 4x^3 \, dx$

$\frac{1}{4} \int \frac{4x^3}{1+(x^4)^2} \, dx$

$\frac{1}{4} \int \frac{du}{1+u^2}$

can you finish?
• Nov 13th 2012, 06:35 PM
NettieL
Re: Integration
Thank you skeeter! Yes, I understand not that x^3 gets replaced by 4x^3dx (please correct me if that is not the correct explanation), and does (u^2)^-1 become just u (later on) because tan^-1 multiplies and flips it somehow? Not sure I have that right...