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Question
Prove the following:
`int_1^3 dx/(x^2(x +1)) = 2/3 + log 2/3`
Solution
Let `I = int_1^3 dx/(x^2 (x + 1))`
Now, `1/(x^2 (x + 1)) = A/x + B/x^2 + C/(x + 1)`
∴ 1 ≡ Ax (x + 1) + B(x + 1) + Cx2 ....(i)
Putting x = 0 in (i), we get
1 = B (0 + 1)
⇒ B = 1
Putting x = -1 in (i), we get
= C (-1)2
⇒ C = 1
Comparing coefficients of x2 on the sides of (i), we get
∴ 0 = A + C
∴ A = - C = - 1
⇒ A = -1
∴ `1/(x^2 (x + 1)) = (- 1)/x + 1/x^2 + 1/(x + 1)`
∴ `int_1^3 1/(x^2 (x + 1))`dx
`= - int_1^3 1/x "dx" + int_1^3 1/x^2 "dx" + int_1^3 1/(x + 1)`dx
`= [- log |x| + x^-1/-1 + log |x + 1|]_1^3`
`= [- 1/x + log |(x + 1)/x|]_1^3 = (-1/3 + 1) + log 4/3 - log 2`
`= 2/3 + log (4/3 xx 1/2)`
`= 2/3 + log 2/3`
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