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Question

The value of the integral 1x41dx is
(where C is an arbitrary constant)

A
14lnx1x+1+12tan1x2+C
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B
12lnx1x+112tan1x+C
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C
12lnx1x+1+14tan1x2+C
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D
14lnx1x+112tan1x+C
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Solution

The correct option is D 14lnx1x+112tan1x+C
1(x41)dx=1(x21)(x2+1)dx=12(1x211x2+1)dx=12(x21)dx12(x2+1)dx=14lnx1x+112tan1x+C

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