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a
−12logx+1+2x+1+1x+1−2x+1+1+C
b
12logx+1−2x+1+1x+1+2x+1+1+C
c
12logx+1+2x+1+1x+1−2x+1+1+C
d
−12logx+1−2x+1+1x+1+2x+1+1+C
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Correct option is B
∫xx2+2x+2x+1dx=∫xx+12+1x+1dx Put x+1=t2dx=2tdt∫t2−12tdtt4+1t=2∫t21−1t2dtt2t2+1t2=2∫1−1t2t2+1t2dt=2∫1−1t2dtt+1t2−(2)2 Put t+1t=k,1−1t2dt=dk=2∫dkk2−(2)2=2122logk−2k+2+C=12logt+1t−2t+1t+2+C=12logt2−2t+1t2+2t+1+C =12logx+1−2x+1+1x+1+2x+1+1+CSimilar Questions
is equal to
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