The correct option is A (x+1)tan−1(2x+23)−34log(4x2+8x+13)+c
I=∫sin−1(2x+2√4x2+8x+13)dx
=∫sin−1⎛⎜
⎜⎝2x+2√(2x+2)2+32⎞⎟
⎟⎠dx
Put 2x+2=3tanθ⇒2dx=3sec2θdθ
I=∫sin−1(3tanθ3secθ)32sec2θdθ
=32∫θsec2θdθ
=32{θtanθ−∫tanθdθ}
=32{θtanθ−log|secθ|}+c
=32⎧⎨⎩2x+23tan−1(2x+23)−log⎛⎝√1+(2x+23)2⎞⎠⎫⎬⎭+c
=32{23(x+1)tan−1(23(x+1))−log√4x2+8x+13}+c
⇒I=(x+1)tan−1(2x+23)−34log(4x2+8x+13)+c