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B
ex√1+xn1−xn+c
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C
−ex1−xn1+xn+c
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D
−ex1+xn1−xn+c
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Solution
The correct option is Bex√1+xn1−xn+c Let I=∫ex(1+nxn−1−x2n)(1−xn)√1−x2ndx =∫ex(1−x2n(1−xn)√1−x2n+nxn−1(1−xn)√1−x2n) Let T=1−x2n(1−xn)√1−x2n=√1−xn1−xn dTdx=nxn−1(1−xn)√1−x2n ∴I=exT=ex√1+xn1−xn