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Q.

I=∫x2−1x32x4−2x2+1dx equal to

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a

2x4−2x2+1+C

b

2x4−2x2+1x3+C

c

2x4−2x2+1x+C

d

2x4−2x2+12x2+C

answer is D.

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Detailed Solution

I=∫1/x3−1/x52−2/x2+1/x4dxput 2−2x2+1x4=t2⇒4x3+4x5dx=2tdt∴   I=12∫tdtt=12t+C=122−2x2+1x4+C=2x4−2x2+12x2+C
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I=∫x2−1x32x4−2x2+1dx equal to