1. ## Integration

Hi there,

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

if u=(x^4)^2

how does the x^3 become 1 in this equation?

2. ## 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?

3. ## 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...