If \int {x^2}/{\sqrt{1-x}} {\d} x= {\p} \sqrt{1-x} (3 x^2+4 x+8) + {\c} where {\c} is a constant of integration, then the value of p is

MHTCET 14 May 2023 Shift 2 Solved Paper

Section: Mathematics

Question : 84 of 150

Marks: +1, -0

If ∫‌

√1−x

dx=p√1−x(3x2+4x+8)+c where c is a constant of integration, then the value of p is

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