Question

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

• A. $\frac{2}{15}$
• B. $\frac{2}{15}$
• C. $\frac{4}{15}$
• D. $-\frac{2}{15}$

Hint:

When evaluating integrals of the form $\int \frac{P(x)}{\sqrt{ax+b dx$, a common strategy is to use the substitution $u = ax+b$ (or $u^2 = ax+b$) to simplify the radical. Always ensure all terms are converted to the new variable before integration.

Answer 1

The Correct Option is A

Step 1: Use substitution.
Let \[ u = 1 - x \Rightarrow x = 1 - u,\quad dx = -du \] \[ \int \frac{x^2}{\sqrt{1-x}}\, dx = \int \frac{(1-u)^2}{\sqrt{u}}(-du) \] \[ = -\int \frac{1 - 2u + u^2}{\sqrt{u}}\, du \] \[ = -\int \left(u^{-1/2} - 2u^{1/2} + u^{3/2}\right) du \]
Step 2: Integrate term-wise.
\[ = -\left[ \frac{u^{1/2}}{1/2} - 2\cdot \frac{u^{3/2}}{3/2} + \frac{u^{5/2}}{5/2} \right] + C \] \[ = -\left[ 2u^{1/2} - \frac{4}{3}u^{3/2} + \frac{2}{5}u^{5/2} \right] + C \]
Step 3: Substitute back $u = 1 - x$.
\[ = -2(1-x)^{1/2} + \frac{4}{3}(1-x)^{3/2} - \frac{2}{5}(1-x)^{5/2} + C \] Factor $2\sqrt{1-x}$: \[ = 2\sqrt{1-x}\left( -1 + \frac{2}{3}(1-x) - \frac{1}{5}(1-x)^2 \right) + C \] Expand: \[ = 2\sqrt{1-x}\left( -1 + \frac{2}{3} - \frac{2}{3}x - \frac{1}{5}(1 - 2x + x^2) \right) + C \] \[ = 2\sqrt{1-x}\left( -1 + \frac{2}{3} - \frac{1}{5} - \frac{2}{3}x + \frac{2}{5}x - \frac{1}{5}x^2 \right) + C \] Combine terms: \[ = 2\sqrt{1-x}\left( -\frac{8}{15} - \frac{4}{15}x - \frac{1}{5}x^2 \right) + C \] \[ = 2\sqrt{1-x}\left( -\frac{3x^2 + 4x + 8}{15} \right) + C \] \[ = -\frac{2}{15}\sqrt{1-x}(3x^2 + 4x + 8) + C \]
Step 4: Compare with given form.
Given form: \[ p\sqrt{1-x}(3x^2 + 4x + 8) + C \] \[ \Rightarrow p = -\frac{2}{15} \]
Final Answer:
\[ \boxed{p = -\frac{2}{15}} \]