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Question

1x4+1dx

A
=122sec1(x212x)122logx22x+1x2+x2+1+C
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B
=122tan1(x212x)142logx22x+1x2+x2+1+C
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C
=142tan1(x212x)142logx22x1x2+x21+C
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D
None of these
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Solution

The correct option is D =122tan1(x212x)142logx22x+1x2+x2+1+C
I=1x4+1dx
=1x2x2+1x2dx
=122x2x2+1x2dx
=12⎜ ⎜ ⎜1+1x2x2+1x211x2x2+1x2⎟ ⎟ ⎟dx
=121+1x2x2+1x2dx1211x2x2+1x2dx
=121+1x2(x1x)2+2dx1211x2(x+1x)22
Putting x1x=u in first integral and x+1x=ν in second integral, we get
I=12duu2+(2)212dvν2(2)2
=122tan1(u2)12×122logν2ν+2+C
=122tan1⎜ ⎜ ⎜x1x2⎟ ⎟ ⎟142log∣ ∣ ∣x+1x2x+1x+2∣ ∣ ∣
=122tan1(x212x)142logx22x+1x2+x2+1+C

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