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Q.

Evaluate ∫1x2−x+1dx

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a

23tan−1⁡x−13+C

b

13tan−1⁡2x−13+C

c

23tan−1⁡2x−13+C

d

-13tan−1⁡2x−13+C

answer is C.

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Detailed Solution

∫1x2−x+1dx=∫1(x−1/2)2+3/4dx=∫1(x−1/2)2+(3/2)2dx=13/2tan−1⁡x−1/23/2+C=23tan−1⁡2x−13+C
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