**Evaluate: **`int (2"x" + 1)/("x"("x - 1")("x - 4"))` dx

#### Solution

Let I = `int (2"x" + 1)/("x"("x - 1")("x - 4"))` dx

Let `(2"x" + 1)/("x"("x - 1")("x - 4")) = "A"/"x" + "B"/"x - 1" + "C"/"x - 4"`

∴ 2x + 1 = A(x - 1)(x - 4) + Bx(x - 4) + Cx(x - 1) ....(i)

Putting x = 0 in (i), we get

0 + 1 = A(0 - 1)(0 - 4) + B(0)(- 4) + C(0)(- 1)

∴ 1 = 4A

∴ A = `1/4`

Putting x = 1 in (i), we get

2(1) + 1 = A(0)(-3) + B(1)(1 - 4) + C(1)(0)

∴ 3 = - 3B

∴ B = - 1

Putting x = 4 in (i), we get

2(4) + 1 = A(3)(0) + B(4)(0) + C(4)(4 - 1)

∴ 9 = C(4)(3)

∴ C = `3/4`

∴ `(2"x" + 1)/("x"("x - 1")("x - 4")) = (1/4)/"x" + (-1)/"x - 1" + (3/4)/"x - 4"`

∴ I = `int((1/4)/"x" + (-1)/("x - 1") + (3/4)/("x - 4"))` dx

`= 1/4 int 1/"x" "dx" - int 1/("x - 1") "dx" + 3/4 int 1/("x - 4")` dx

∴ I = `1/4 log |"x"| - log |"x - 1"| + 3/4 log |"x - 4"| + "c"`