Use partial fractions to calculate the following integrals: 1 ∫ (x+3)(x-4)/(2x³-x²-6x) dx 2 ∫ (3x²+16x+26)/((x-2)(x²+3x+4)) dx

The numerator is $(x+3)(x-4) = x^2 - x - 12$.
The denominator is $2x^3 - x^2 - 6x = x(2x^2 - x - 6)$.
Factor the quadratic $2x^2 - x - 6$:
$2x^2 - x - 6 = 2x^2 - 4x + 3x - 6 = 2x(x-2) + 3(x-2) = (2x+3)(x-2)$.
So the integrand is $\frac{x^2 - x - 12}{x(2x+3)(x-2)}$.

Multiply both sides by $x(2x+3)(x-2)$:

• Set $x=0$: $0^2 - 0 - 12 = A(2(0)+3)(0-2) + B(0)(0-2) + C(0)(2(0)+3)$ $-12 = A(3)(-2) \implies -12 = -6A \implies A = 2$.
• Set $x=2$: $2^2 - 2 - 12 = A(2(2)+3)(2-2) + B(2)(2-2) + C(2)(2(2)+3)$ $4 - 2 - 12 = 0 + 0 + C(2)(7) \implies -10 = 14C \implies C = -\frac{10}{14} = -\frac{5}{7}$.
• Set $x = -\frac{3}{2}$: $\left(-\frac{3}{2}\right)^2 - \left(-\frac{3}{2}\right) - 12 = A(\dots) + B\left(-\frac{3}{2}\right)\left(-\frac{3}{2}-2\right) + C(\dots)$ $\frac{9}{4} + \frac{3}{2} - 12 = B\left(-\frac{3}{2}\right)\left(-\frac{7}{2}\right)$ $\frac{9+6-48}{4} = B\left(\frac{21}{4}\right)$ $-\frac{33}{4} = \frac{21}{4}B \implies B = -\frac{33}{21} = -\frac{11}{7}$.

For the quadratic factor $x^2 + 3x + 4$, the discriminant is $\Delta = b^2 - 4ac = 3^2 - 4(1)(4) = 9 - 16 = -7$.
Since the discriminant is negative, $x^2 + 3x + 4$ is irreducible over real numbers.

Multiply both sides by $(x-2)(x^2+3x+4)$:

• Set $x=2$: $3(2)^2 + 16(2) + 26 = A(2^2+3(2)+4) + (B(2)+C)(2-2)$ $12 + 32 + 26 = A(4+6+4) + 0$ $70 = 14A \implies A = 5$.

• Expand the right side and equate coefficients: $3x^2 + 16x + 26 = Ax^2+3Ax+4A + Bx^2 - 2Bx + Cx - 2C$ $3x^2 + 16x + 26 = (A+B)x^2 + (3A-2B+C)x + (4A-2C)$

  Equating coefficients of $x^2$:
  $A+B = 3 \implies 5+B=3 \implies B = -2$.

  Equating constant terms:
  $4A-2C = 26 \implies 4(5)-2C=26 \implies 20-2C=26 \implies -2C=6 \implies C = -3$.
  (To verify, check the coefficient of $x$: $3A-2B+C = 3(5)-2(-2)+(-3) = 15+4-3 = 16$. This matches the left side.)

The first integral is $5 \ln|x-2|$.
For the second integral, $I_2 = \int \frac{2x+3}{x^2+3x+4} dx$:
Notice that the numerator $2x+3$ is the derivative of the denominator $x^2+3x+4$.
Let $u = x^2+3x+4$, then $du = (2x+3)dx$.

(Since $x^2+3x+4$ has a negative discriminant and a positive leading coefficient, it is always positive, so absolute value is not strictly needed).