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: Understanding the Concept
Treat \(x^2\) as a single variable and write the integrand as \(\dfrac{A}{x^2 + 2} + \dfrac{B}{x^2 + 5}\).

Step 2: Find A and B
\[ x^2 = A(x^2 + 5) + B(x^2 + 2) \]
Comparing coefficients: \(A + B = 1\) and \(5A + 2B = 0\). Solving gives \(A = -\dfrac23\) and \(B = \dfrac53\).

Step 3: Integrate
Using \(\int\dfrac{dx}{x^2 + a^2} = \dfrac1a\tan^{-1}\dfrac xa\):
\[ -\frac23\cdot\frac{1}{\sqrt2}\tan^{-1}\frac{x}{\sqrt2} + \frac53\cdot\frac{1}{\sqrt5}\tan^{-1}\frac{x}{\sqrt5} + c \]
\[ = -\frac{\sqrt2}{3}\tan^{-1}\frac{x}{\sqrt2} + \frac{\sqrt5}{3}\tan^{-1}\frac{x}{\sqrt5} + c \]
This is option (D). The signs matter: the \(\sqrt2\) term is negative and the \(\sqrt5\) term is positive.

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