Question

\(∫\frac{dx}{(x2+1) (x2+4)} =\)

• A. \(\frac{1}{3}Tan^{-1}x +\frac{1}{6}Tan^{-1}(\frac{x}{2})+c\)
• B. \(\frac{1}{3}Tan^{-1}x -\frac{1}{3}Tan^{-1}(\frac{x}{2})+c\)
• C. \(\frac{1}{3}Tan^{-1}x +\frac{1}{3}Tan^{-1}(\frac{x}{2})+c\)
• D. \(\frac{1}{3}Tan^{-1}x -\frac{1}{6}Tan^{-1}(\frac{x}{2})+c\)

Answer 1

The Correct Option is D

To solve the integral \(∫\frac{dx}{(x^2+1) (x^2+4)}\), we need to use the method of partial fraction decomposition. This technique is useful for integrating rational functions.

Step 1: Partial Fraction Decomposition

We express the integrand \(\frac{1}{(x^2+1)(x^2+4)}\) as a sum of partial fractions:

\(\frac{1}{(x^2+1)(x^2+4)} = \frac{Ax + B}{x^2+1} + \frac{Cx + D}{x^2+4}\)

Step 2: Solve for Coefficients

Equating the numerators, we have:

\(1 = (Ax + B)(x^2 + 4) + (Cx + D)(x^2 + 1)\)

Expanding both terms:

\((Ax^3 + 4Ax + Bx^2 + 4B) + (Cx^3 + Dx^2 + Cx + D) = 1\)

Combine like terms:

\(x^3(A + C) + x^2(B + D) + x(4A + C) + (4B + D) = 1\)

Comparing coefficients with the polynomial \(0x^3 + 0x^2 + 0x + 1\), we get the system of equations:

• \(A + C = 0\)
• \(B + D = 0\)
• \(4A + C = 0\)
• \(4B + D = 1\)

Solving, we find: \(A = 0\), \(C = 0\), \(B = \frac{1}{3}\), \(D = -\frac{1}{3}\).

Step 3: Integrate Each Term

The partial fractions are:

\(\frac{1}{3} \cdot \frac{1}{x^2+1} - \frac{1}{3} \cdot \frac{1}{x^2+4}\)

Integrating each term separately, using the formula \(\int \frac{1}{x^2+a^2} \, dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right)\), gives:

• \(\frac{1}{3} \int \frac{1}{x^2+1} \, dx = \frac{1}{3} \tan^{-1} x\)
• \(-\frac{1}{3} \int \frac{1}{x^2+4} \, dx = -\frac{1}{3} \cdot \frac{1}{2} \tan^{-1} \left(\frac{x}{2}\right)\)
• This simplifies to -\frac{1}{6} \tan^{-1} \left(\frac{x}{2}\right)

Conclusion

Combining the results, the integral is:

\(\frac{1}{3} \tan^{-1}x - \frac{1}{6} \tan^{-1} \left(\frac{x}{2}\right) + c\)

The correct answer is:

\(\frac{1}{3}Tan^{-1}x -\frac{1}{6}Tan^{-1}(\frac{x}{2})+c\)