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Sum

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

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#### Solution

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

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

∴ 2x + 1 = A(x - 2) + B(x + 1) ....(i)

Putting x = - 1 in (i), we get

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

∴ - 1 = -3A

∴ A = `1/3`

Putting x = 2 in (i), we get

2(2) + 1 = A(0) + B(3)

∴ 5 = 3B

∴ B = `5/3`

∴ `(2"x" + 1)/(("x + 1")("x - 2")) = (1/3)/"x + 1" + (5/3)/"x - 2"`

∴ I = `int (((1/3))/"x + 1" + ((5/3))/"x - 2")` dx

∴ `1/3 int 1/"x + 1" "dx" + 5/3 int 1/"x - 2"` dx

∴ I = `1/3 log |"x" + 1| + 5/3 log |"x - 2"| + "c"`

Concept: Methods of Integration: Integration Using Partial Fractions

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