Question

\[ \int \frac{x^3}{x^4+5 x^2+4} d x= \]

• A. \(\frac{1}{3}\mathrm{log}(\frac{(x^{2}+4)^{2}}{\sqrt{x^{2}+1}})+\mathrm{c}\), where c is the constant of integration
• B. \(\mathrm{log}(\frac{(x^{2}+4)^{2}}{\sqrt{x^{2}+1}})+\mathrm{c}\), where c is the constant of integration
• C. \(3\mathrm{log}(\frac{(x^{2}+4)^{2}}{\sqrt{x^{2}+1}})+\mathrm{c}\), where c is the constant of integration
• D. \(\frac{2}{3}\mathrm{log}(\frac{(x^{2}+4)^{2}}{\sqrt{x^{2}+1}})+\mathrm{c}\), where c is the constant of integration

Hint:

Put u = x^2 and split the resulting fraction into partial fractions.

Answer 1

The Correct Option is A

Step 1: Factorise
$x^{4}+5x^{2}+4=(x^{2}+1)(x^{2}+4)$.

Step 2: Split
Write $\dfrac{x^{3}}{(x^{2}+1)(x^{2}+4)}=\dfrac{Ax}{x^{2}+1}+\dfrac{Bx}{x^{2}+4}$. This gives $A+B=1$ and $4A+B=0$, so $A=-\dfrac13,\ B=\dfrac43$.

Step 3: Integrate
$\int\dfrac{x}{x^{2}+k}dx=\dfrac12\log(x^{2}+k)$, so $I=-\dfrac16\log(x^{2}+1)+\dfrac23\log(x^{2}+4)+c=\dfrac13\log\dfrac{(x^{2}+4)^{2}}{\sqrt{x^{2}+1}}+c$.

Final Answer:
Direct partial fractions give the same logarithmic expression. \[ \boxed{\text{(A) }\dfrac{1}{3}\log\left(\dfrac{(x^{2}+4)^{2}}{\sqrt{x^{2}+1}}\right)+c} \]