Integration and its Applications — Important Questions

1/((x-1)(x+2)) = A/(x-1) + B/(x+2), x/((x-1)²) = A/(x-1) + B/(x-1)², (x+1)/((x-2)(x+3)) = A/(x-2) + B/(x+3)

Options 1, 2, and 3 follow correct partial fraction rules. Option 4 is incorrect because x²+1 cannot be factored into linear factors with real coefficients.

e^x(x-1) + C

Using ∫udv = uv - ∫vdu with u=x, dv=e^x dx: ∫xe^x dx = xe^x - ∫e^x dx = xe^x - e^x = e^x(x-1) + C.

log|x| + C

The integral of 1/x is log|x| + C. We use absolute value because x can be negative, and the logarithm of a negative number is undefined in real numbers.