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Question

# If we convert the denominator of the integral into a perfect square, ∫1x2 − x + 1dx then the correct integral will be

A

1(x 1/2)2 + 1dx

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B

1(x 1/2)2 + 3/4dx

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C

1(x 1)2 + 1/2dx

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D

None of these

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Solution

## The correct option is B ∫1(x − 1/2)2 + 3/4dx We know that ax2 + bx + c can also be written as a [(x + b/2a)2 + c/a − b2/4a2] by completing squares, which we learnt in quadratic equations. We'll apply the same formula here. In the question given here, a = 1, b = -1, c = 1 So, x2 − x + 1 = 1[(x − 1/2)2 + 1 − 1/4] or x2 − x + 1 = [(x − 1/2)2 + 3/4] The given integral ∫1x2 − x + 1dx can be written as ∫1(x − 1/2)2 + 3/4dx

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