Question

solve this..............

Answer 1

Hi there!
Here's the answer:

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(x-1)/[(x-2)(x²+1)]
= (x-1)/[(x-2)(x+1)(x-1)]
= 1/[(x-2)(x+1)]

Now Resolve into Partial Fractions

1/[(x-2)(x+1)]
= [A/(x-2)] + [B/(x+1)]
……………_____________(1)

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

Separate constants and variables
=> x(A+B) + (A-2B) = 1

Compare coefficients
=> A+B = 0 => A= -B
& A-2B = 1

substitute A= -B
=> -B-2B = 1
=> -3B = 1
=> B = -(1/3)

°•°A = -B = -(-1/3) = (1/3)

•°• A = (1/3) & B = -(1/3)

Substitute in (1)
=> (1/3)[1/(x-2)] + (-1/3)[1/(x+1)]
=> [1/3] { [1/(x-2)] - [1/(x+1)] }

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Now,

∫ (x-1)/[(x-2)(x²+1)] dx
= ∫ [1/3] { [1/(x-2)] - [1/(x+1)] } dx

Apply Integration
=> (1/3) [ log(x-2) - log(x+1)]

°•° ∫ (1/x) dx = log x

°•° log a - log b = log (a/b)

= (1/3)× log [(x-2)/(x+1)]

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Hope it helps