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: Understanding the Concept
The numerator \(x^{3}dx\) can be written with \(u=x^{2}\): \(x^{3}dx=\dfrac{u\,du}{2}\).

Step 2: Substitute
\(x^{4}+5x^{2}+4=u^{2}+5u+4=(u+1)(u+4)\), so \(I=\dfrac12\int\dfrac{u\,du}{(u+1)(u+4)}\).

Step 3: Partial fractions
\(\dfrac{u}{(u+1)(u+4)}=\dfrac{-1/3}{u+1}+\dfrac{4/3}{u+4}\).

Step 4: Integrate
\(I=\dfrac12\left[-\dfrac13\log(u+1)+\dfrac43\log(u+4)\right]=\dfrac16\left[4\log(x^{2}+4)-\log(x^{2}+1)\right]+c\).

Step 5: Rewrite
\(=\dfrac13\log\dfrac{(x^{2}+4)^{2}}{\sqrt{x^{2}+1}}+c\).

Final Answer:
The integral is (1/3) log((x^2 + 4)^2 / sqrt(x^2 + 1)) + c. \[ \boxed{\text{(A) }\dfrac{1}{3}\log\left(\dfrac{(x^{2}+4)^{2}}{\sqrt{x^{2}+1}}\right)+c} \]