Question

If $\int \frac{x^2}{\sqrt{1-x}}dx=p\sqrt{1-x}\left(3x^2+4x+8\right)$, then the value of $p$ is:

• A. $-\frac{2}{15}$
• B. $\frac{2}{15}$
• C. $\frac{4}{15}$
• D. None of these

Answer 1

The correct option is A

$-\frac{2}{15}$

Explanation for the correct option:

Integration:

$\int \frac{x^2}{\sqrt{1-x}}dx$

Let $\begin{array}{rcl}1-x& =& t^2\end{array}$ then $x=1-t^{2}anddx=-2tdt$. So,

$\begin{array}{rcl}\int \frac{x^2}{\sqrt{1-x}}dx& =& -2\int \frac{t{\left(1-t^2\right)}^2}{t}dt\\ & =& -2\int \left(1-2t^2+t^4\right)dt\\ & =& -2\left(t-\frac{2t^3}{3}+\frac{t^5}{5}\right)+c\\ & =& -2\left(\frac{15t-10t^3+3t^5}{15}\right)+c\\ & =& -2t\left(\frac{15-10t^2+3t^4}{15}\right)+c\end{array}$

By substituting the value of $t$, we get

$$\begin{gathered} =-2\left(\sqrt{1-x}\right)\left(\frac{15-10\left(1-x\right)+3{\left(1-x\right)}^2}{15}\right)+c \\ =-2\left(\sqrt{1-x}\right)\left(\frac{15-10+10x+3+3x^2-6x}{15}\right)+c \\ =\frac{-2\left(\sqrt{1-x}\right)}{15}\left(3x^2+4x+8\right) \end{gathered}$$

So, $p=\frac{-2}{15}$

Hence, option A is correct.