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Question

Evaluate: xddx(sec1x) dx for x>1

A
ln(x21+x2)+C
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B
ln(x21x)+C
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C
ln(x21+x)x+C
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D
ln(x21+x)+C
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Solution

The correct option is D ln(x21+x)+C
x.d(sec1x)dxdx=x.1xx21dx=dxx21=logx+x21+C

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