Question

$\int \frac{x^3}{x^4+5x^2+4} dx =$

• A. $\frac{1}{3} \log \left( \frac{(x^2+4)^2}{\sqrt{x^2+1}} \right) + c$
• B. $\log \left( \frac{(x^2+4)^2}{\sqrt{x^2+1}} \right) + c$
• C. $3 \log \left( \frac{(x^2+4)^2}{\sqrt{x^2+1}} \right) + c$
• D. $\frac{2}{3} \log \left( \frac{(x^2+4)^2}{\sqrt{x^2+1}} \right) + c$

Hint:

Use substitution $x^2=t$ when the numerator has $x$ or $x^3$ and the denominator is in terms of $x^2$.

Answer 1

The Correct Option is D

Step 1: Substitution
Let $x^2 = t \implies 2x dx = dt \implies x dx = \frac{1}{2} dt$.
Integral becomes $\frac{1}{2} \int \frac{t}{t^2 + 5t + 4} dt = \frac{1}{2} \int \frac{t}{(t+4)(t+1)} dt$.
Step 2: Partial Fractions
$\frac{t}{(t+4)(t+1)} = \frac{4/3}{t+4} - \frac{1/3}{t+1}$.
$\frac{1}{2} \int \left( \frac{4/3}{t+4} - \frac{1/3}{t+1} \right) dt = \frac{2}{3} \log(t+4) - \frac{1}{6} \log(t+1)$.
Step 3: Re-substitute
$\frac{1}{6} [4 \log(x^2+4) - \log(x^2+1)] = \frac{1}{6} \log \frac{(x^2+4)^4}{x^2+1} = \frac{1}{3} \log \frac{(x^2+4)^2}{\sqrt{x^2+1}}$.
Checking options, (A) matches the simplified form $1/3 .......$ let's verify coefficients. $4/6 = 2/3$. Thus (A) or (D) depend on final simplification.
Final Answer: (A)