Question

If \[ \int \frac{x^3}{\sqrt{1 + x^2}} \, dx = A(1 + x^2)^{\frac{3}{2}} + B(1 + x^2)^{\frac{1}{2}} + C, \text{ then } A + B = \]

• A. \( \frac{2}{3} \)
• B. \( -\frac{2}{3} \)
• C. \( \frac{1}{3} \)
• D. \( -\frac{1}{3} \)

Hint:

To identify constants in an integral expression, differentiate both sides and compare coefficients.

Answer 1

The Correct Option is B

We are given that: \[ \int \frac{x^3}{\sqrt{1 + x^2}} \, dx = A(1 + x^2)^{\frac{3}{2}} + B(1 + x^2)^{\frac{1}{2}} + C \]
Differentiate both sides with respect to \( x \) using the chain rule: \[ \frac{x^3}{\sqrt{1 + x^2}} = \frac{d}{dx}\left[ A(1 + x^2)^{\frac{3}{2}} + B(1 + x^2)^{\frac{1}{2}} \right] \]
\[ = A \cdot \frac{3}{2}(1 + x^2)^{\frac{1}{2}} \cdot 2x + B \cdot \frac{1}{2}(1 + x^2)^{-\frac{1}{2}} \cdot 2x \]
\[ = 3Ax(1 + x^2)^{\frac{1}{2}} + Bx(1 + x^2)^{-\frac{1}{2}} \]
Multiply both terms by \( (1 + x^2)^{\frac{1}{2}} \) to simplify the left-hand side: \[ x^3 = 3A x (1 + x^2) + Bx \Rightarrow x^3 = x \left[ 3A(1 + x^2) + B \right] \]
Divide both sides by \( x \) (since \( x \neq 0 \)): \[ x^2 = 3A(1 + x^2) + B \Rightarrow x^2 = 3A + 3A x^2 + B \]
Rearranging: \[ x^2 - 3A x^2 = 3A + B \Rightarrow x^2(1 - 3A) = 3A + B \]
Now equating coefficients: - Coefficient of \( x^2 \): \( 1 - 3A = 0 \Rightarrow A = \frac{1}{3} \)
- Constant term: \( 3A + B = 0 \Rightarrow B = -3A = -1 \)
\[ A + B = \frac{1}{3} - 1 = -\frac{2}{3} \]