Evaluate: ∫x²/((x-1)²(x+3))dx

∫x²/((x-1)²(x+3))dx = ∫(x-1+1)/((x-1)²(x+3))dx = ∫(x-1)/((x-1)²(x+3))dx + ∫1/((x-1)²(x+3))dx = ∫1/((x-1)(x+3))dx + ∫1/((x-1)²(x+3))dx
Let's solve the first integral:
∫1/((x-1)(x+3))dx = ∫(1/4)(1/(x-1) - 1/(x+3))dx = (1/4)[ln|x-1| - ln|x+3|] + c₁ = (1/4)ln|(x-1)/(x+3)| + c₁
Now let's solve the second integral using partial fraction decomposition:
1/((x-1)²(x+3)) = A/(x-1) + B/(x-1)² + C/(x+3)
1 = A(x-1)(x+3) + B(x+3) + C(x-1)²
If x = 1, 1 = 4B => B = 1/4
If x = -3, 1 = 16C => C = 1/16
If x = 0, 1 = -3A + 3B + C => 1 = -3A + 3(1/4) + 1/16 => 1 = -3A + 13/16 => 3A = -3/16 => A = -1/16
So, ∫1/((x-1)²(x+3))dx = ∫(-1/16)/(x-1)dx + ∫(1/4)/(x-1)²dx + ∫(1/16)/(x+3)dx
= (-1/16)ln|x-1| - (1/4)/(x-1) + (1/16)ln|x+3| + c₂
Combining the results:
∫x²/((x-1)²(x+3))dx = (1/4)ln|(x-1)/(x+3)| + (-1/16)ln|x-1| - 1/(4(x-1)) + (1/16)ln|x+3| + c
= (1/16)[4ln|(x-1)/(x+3)| - ln|x-1| + ln|x+3|] - 1/(4(x-1)) + c
= (1/16)[ln|((x-1)/ (x+3))^4| + ln|(x+3)/(x-1)|] - 1/(4(x-1)) + c
= (1/16)ln|((x-1)/(x+3))^3/(x-1)| - 1/(4(x-1)) + c
= (1/16)ln| (x-1)²/(x+3)³| - 1/(4(x-1)) + c