Question

\(\int \frac{x^2\,\text{d}x}{(x^2+2)(x^2+5)} =\)

• A. \(-(\frac{\sqrt{2}}{3}tan^{-1}\frac{x}{\sqrt{2}}+\frac{\sqrt{5}}{3}tan^{-1}\frac{x}{\sqrt{5}})+c\), where c is the constant of integration
• B. \((\frac{\sqrt{2}}{3}tan^{-1}\frac{x}{\sqrt{2}}+\frac{\sqrt{5}}{3}tan^{-1}\frac{x}{\sqrt{5}})+c\), where c is the constant of integration
• C. \(\frac{\sqrt{2}}{3}tan^{-1}(\frac{x}{\sqrt{2}})-\frac{\sqrt{5}}{3}tan^{-1}(\frac{x}{\sqrt{5}})+c\), where c is the constant of integration
• D. \(-\frac{\sqrt{2}}{3}tan^{-1}(\frac{x}{\sqrt{2}})+\frac{\sqrt{5}}{3}tan^{-1}(\frac{x}{\sqrt{5}})+c\), where c is the constant of integration

Hint:

Split into partial fractions in x^2 and integrate each as an inverse tangent.

Answer 1

The Correct Option is D

Step 1: Rewrite numerator
$x^2 = \dfrac{5(x^2 + 2) - 2(x^2 + 5)}{3}$ since $5x^2 + 10 - 2x^2 - 10 = 3x^2$.

Step 2: Split
So the integrand is $\dfrac{5}{3(x^2 + 5)} - \dfrac{2}{3(x^2 + 2)}$.

Step 3: Integrate
This gives $\dfrac{\sqrt5}{3}\tan^{-1}\dfrac{x}{\sqrt5} - \dfrac{\sqrt2}{3}\tan^{-1}\dfrac{x}{\sqrt2} + c$. Option (D).

Final Answer:
Option (D). \[ \boxed{-\frac{\sqrt{2}}{3}\tan^{-1}\frac{x}{\sqrt{2}} + \frac{\sqrt{5}}{3}\tan^{-1}\frac{x}{\sqrt{5}} + c} \]