Asked by Mike
Solve the integral: ∫ x^3/(x^+2)^2
I'm stuck. This is my work so far.
∫ A/(x^+2) + ∫ B/(x^2-2)^2
F(x)=A ln(x^2+2)-B/(x^2+2)+C
X^3=A(x^2+2)+B
I'm stuck. This is my work so far.
∫ A/(x^+2) + ∫ B/(x^2-2)^2
F(x)=A ln(x^2+2)-B/(x^2+2)+C
X^3=A(x^2+2)+B
Answers
Answered by
Damon
let z = x^2+2 then dz = 2x dx and dx = dz/2x and x = (z-2)^.5 and xdx=dz/2
so x^2 = (z-2)
now what you have is
∫ (z-2) x dx/z^2 = ∫ (z-2)dz /2z^2
= ∫ dz/2z - ∫ dz/z^2
so x^2 = (z-2)
now what you have is
∫ (z-2) x dx/z^2 = ∫ (z-2)dz /2z^2
= ∫ dz/2z - ∫ dz/z^2
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