If 4∑i=1(sin−1xi+cos−1yi)=6π, then 4∑i=1yi∫4∑i=1xi(−1+l – InterviewSolution

1.	If 4∑i=1(sin−1xi+cos−1yi)=6π, then 4∑i=1yi∫4∑i=1xi(−1+ln(x+√x2+1)+5x3+2x2)ex−2dx=AeB. The value of AB is
Answer» If $4 \sum i = 1 ({sin}^{- 1} x_{i} + {cos}^{- 1} y_{i}) = 6 π$ , then $4 \sum i = 1 y_{i} \int 4 \sum i = 1 x_{i} (- 1 + ln (x + \sqrt{x^{2} + 1}) + 5 x^{3} + \frac{2}{x^{2}}) e^{x^{- 2}} d x = A e^{B}$ . The value of $\frac{A}{B}$ is

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