find the volume generated by revolving the area bounded by
y=1/(x^3+15x^2+56x), x=2, x=5, and y=0 about the y-axis.
AAAaannndd the bot gets it wrong yet again!
using shells of thickness dx,
v = ∫[2,5] 2πrh dx
where r=x and h=y
v = ∫[2,5] 2πx/(x^3+15x^2+56x) dx
= ∫[2,5] 2π/(x^2+15x+56) dx = 2πln(40/39) = 0.159
solve that using partial fractions:
1/(x^2+15x+6) = 1/((x+7)(x+8) = 1/(x+7) - 1/(x+8)
so
∫1/(x^2+15x+6) dx = ln(x+7) - ln(x+8) = ln (x+7)/(x+8)