If ∫(x2 / √1 - x) dx = p √(1 - x) [3x2 + 4x + 8], then the value of p is

1) – 2 / 15

2) 2 / 15

3) 4 / 15

4) None of these

Solution: (1) – 2 / 15

\(\begin{array}{l}\begin{array}{l} \int \frac{x^{2} d x}{\sqrt{1-x}}=-\int \frac{\left(1-z^{2}\right)^{2}}{z} 2 z d z \\ \text { Let } 1-x=z^{2} \\ -d x=2 z d z \\ =-2 \int\left(1-z^{2}\right)^{2} d z \\ =\int\left(1+z^{4}-2 z^{2}\right) d z \\ =-2\left[z+\frac{z^{5}}{5}-\frac{2 z^{3}}{3}\right]+c \\ =-2 z\left[1+\frac{z^{4}}{5}-\frac{2 z^{2}}{3}\right]+c \\ =-2 \sqrt{1-x}\left[1+\frac{(1-x)^{2}}{5}-\frac{2(1-x)}{3}\right]+c \\ =-2 \sqrt{1-x}\left[1+\frac{1+x^{2}-2 x}{5}-\frac{2-2 x}{3}\right]+c \\ =-\frac{2}{15} \sqrt{1-x}\left[15+3+3 x^{2}-6 x-10+10 x\right]+c \\ =-\frac{2}{15} \sqrt{1-x}\left[3 x^{2}+4 x+8\right] \\ P=-\frac{2}{15} \end{array}\end{array} \)

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