Verified answer

We need to express as partial fractions.

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

=> x + 1 = A(x + 2) + B(x - 1)

If x = 1 we get:

2 = 3A => A = 2/3

If x = -2 we get:

-1 = -3B => B = 1/3

i.e. (x + 1)/(x - 1)(x + 2) => 2/3(x - 1) + 1/3(x + 2)

Then, ∫ [(x + 1)/(x - 1)(x + 2)] dx = ∫ [2/3(x - 1) + 1/3(x + 2)] dx

Therefore, (2/3)ln(x - 1) + (1/3)ln(x + 2) + C

or. (1/3)ln[(x - 1)²(x + 2)] + C

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∫ (x+1)/((x-1)(x+2)) dx

= ∫ A/(x-1) dx + ∫ B/(x+2) dx

= ∫ (A(x+2) + B(x-1))dx/((x-1)(x+2))

for x = 1 we get for A is:

(x + 1) = A(x + 2) + B(x - 1)

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

2 = 3A + 0

A = 2/3

for x = -2 we get for B is:

(x + 1) = A(x + 2) + B(x - 1)

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

-1 = 0 - 3B

B = 1/3

Now, change A with 2/3 and change B with 1/3 on ∫ A/(x-1) dx + B/(x+2) dx, we get:

∫ A/(x-1) dx + ∫ B/(x+2) dx

= ∫ (2/3)/(x-1) dx + ∫ (1/3)/(x+2) dx

= 2/3 ∫ dx/(x - 1) + 1/3 ∫ dx/(x + 2)

= 2/3 ln |x - 1| + 1/3 ln |x + 2| + c

∫ (x+1)/(x-1)(x+2) dx

Decompose (x+1)/(x-1)(x+2) into partial fractions.

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

Multiply everything by (x-1)(x+2)

(x+1) = A(x+2) + B(x-1)

Equate the coefficient of x on both sides

1 = A+B ---- (1)

Equate the constant terms from both sides

1 = 2A - B ---(2)

Add (1) & (2)

3A=2

A=2/3

B = 1-2/3 = 1/3

∫ (x+1)/(x-1)(x+2) dx = ∫ [A /(x-1) + B / (x+2)] dx = (2/3) ∫ dx /(x-1) + (1/3) ∫ dx /(x+2)

= (2/3) ln(x-1) + (1/3) ln(x+2) + C