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B
x+1x+2
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C
x−1x+1
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D
x−1x−1
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Solution
The correct option is Ax+1x+1 I=∫(x−1)dx(x+1)x√x+1+1x=∫(x−1)(x+1)dx(x+1)2x√x+1+1x=∫(1−1x2)dx(x+1x+2)√x+1x+1Putx+1+1x=t2(1−1x2)dx=2tdt=∫2tdt(t2+1)t=2tan−1t+c=2tan−1(√x+1x+1)+c
f(x) = x+1x+1