Given ![]()
Expressing the integral ![]()
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Consider ![]()
By partial fraction decomposition,
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⇒ 2x + 1 = Ax + B(x – 1)
⇒ 2x + 1 = Ax + Bx – B
⇒ 2x + 1 = (A + B)x – B
∴ B = -1 and A + B = 2
∴ A = 2 + 1 = 3
Thus, ![]()
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Consider ![]()
Substitute u = x – 1 → dx = du.
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We know that ![]()
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Then,
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∴ ![]()
Then,
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We know that ![]()
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∴ ![]()
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