Question

\(\int \frac{x^3}{x^4 + 3x^2 + 2} dx =\)

• A. \(\log \left( \frac{x^2+2}{\sqrt{x^2+1}} \right) + c\)
• B. \(\log(x^2+2) - 2\log(x^2+1) + c\)
• C. \(\log \left( \frac{(x^2+2)x}{\sqrt{x^2+1}} \right) + c\)
• D. \(\log \left( \frac{x^2+1}{\sqrt{x^2+2}} \right) + c\)

Hint:

When the integrand involves only even powers of \(x\) and an \(x\) or \(x^3\) in the numerator, substituting \(t=x^2\) is the standard approach to simplify the degree.

Answer 1

The Correct Option is A

Step 1: Substitution: Let \(x^2 = t\). Then \(2x dx = dt \implies x dx = \frac{dt}{2}\). Rewrite the integrand: \[ \int \frac{x^2 \cdot x}{x^4 + 3x^2 + 2} dx = \int \frac{t}{t^2 + 3t + 2} \frac{dt}{2} \] \[ = \frac{1}{2} \int \frac{t}{(t+1)(t+2)} dt \]
Step 2: Partial Fractions: Let \(\frac{t}{(t+1)(t+2)} = \frac{A}{t+1} + \frac{B}{t+2}\). \(t = A(t+2) + B(t+1)\). At \(t = -1\): \(-1 = A(1) \implies A = -1\). At \(t = -2\): \(-2 = B(-1) \implies B = 2\).
Step 3: Integration: \[ I = \frac{1}{2} \left[ \int \frac{-1}{t+1} dt + \int \frac{2}{t+2} dt \right] \] \[ I = \frac{1}{2} [ -\log(t+1) + 2\log(t+2) ] + c \] \[ I = \frac{1}{2} [ \log \frac{(t+2)^2}{t+1} ] + c \] \[ I = \log \left( \sqrt{\frac{(t+2)^2}{t+1}} \right) + c = \log \left( \frac{t+2}{\sqrt{t+1}} \right) + c \]
Step 4: Back Substitution: Replace \(t\) with \(x^2\): \[ I = \log \left( \frac{x^2+2}{\sqrt{x^2+1}} \right) + c \]