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Question

Evaluate sin1(2x+24x2+8x+13) dx

A
(x+1)tan1(2x+23)34log(4x2+8x+13)+c
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B
(x+1)tan1(23)34log(x2+8x+13)+c
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C
tan1(2x+23)+34log(4x2+8x+13)+c
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D
none of these
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Solution

The correct option is A (x+1)tan1(2x+23)34log(4x2+8x+13)+c
I=sin1(2x+24x2+8x+13)dx
=sin1⎜ ⎜2x+2(2x+2)2+32⎟ ⎟dx

Put 2x+2=3tanθ2dx=3sec2θdθ
I=sin1(3tanθ3secθ)32sec2θdθ
=32θsec2θdθ
=32{θtanθtanθdθ}
=32{θtanθlog|secθ|}+c
=322x+23tan1(2x+23)log1+(2x+23)2+c
=32{23(x+1)tan1(23(x+1))log4x2+8x+13}+c
I=(x+1)tan1(2x+23)34log(4x2+8x+13)+c

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