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Question

The value of x2+1x4x2+1dx is

A
tan1(2x21)+C
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B
tan1(x21x)+C
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C
sin1(x1x)+C
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D
tan1x2+C
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Solution

The correct option is A tan1(x21x)+C
Let I=x2+1x4x2+1dx=1+1x2x2+1x21dx
=(1+1x2)(x1x)2+1dx
Put x1x=t(1+1x2)dx=dt
I=dtt2+1=tan1t+C=tan1(x1x)+C
=tan1(x21x)+C

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