Asked by Emily

Find the integral

∫(x^2-6x+1)/(x^2+1) dx
Answered by Damon
∫(x^2-6x+1)/(x^2+1) dx

=∫(x^2+1)dx/( x^2+1)-6∫xdx/( x^2+1)

= ∫dx - 6∫xdx/( x^2+1)
Answered by Damon
x - 6∫xdx/( x^2+1)

let z = x^2+1
dz = 2 x dx
so x dx = (1/2)dz
then
x - 6∫(1/2)dz/z

x - 3 ln z

x - 3 ln(x^2+1)
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