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Question

tan1xdx=

A
xtan1x12log|1+x2|+c
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B
xtan1x+12log|1+x2|+c
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C
xtan1x12log|1+x|+c
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D
xtan1x12log|1+x3|+c
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Solution

The correct option is A xtan1x12log|1+x2|+c
I=tan1xI×1IIdx
From integration by parts,
I=tan1xx11+x2xdx
I1=xx2+1dx Let x2+1=t
2xdx=dt
=dt2tI1=12lnt+c=12ln|x2+1|+c
I=xtan1x12ln|x2+1|+c
(A).

1181965_1291183_ans_01c10640d909428ea5e9fb10db6092a3.jpg

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