Definite Integration
Indefinite Integration
Grade Class 12
Question:
∫ \frac{x^2 + 3}{x^6(x^2 + 1)} dx equals
C - \frac{2}{x} + \frac{2}{3x^3} - \frac{3}{5x^5} - 2\tan^{-1} x
C - \frac{2}{x^2} + \frac{2}{3x^4} - \frac{3}{5x^6} - 2\tan^{-1} x
C - \frac{2}{x} + \frac{2}{3x^3} - \frac{3}{5x^5} + 2\tan^{-1} x
C - \frac{2}{x^2} + \frac{2}{3x^4} + \frac{3}{5x^6} + 2\tan^{-1} x
Step-by-Step Solution
Key Concept: The integral can be solved by expressing the integrand as (x^2+1+2)/(x^6(x^2+1)) = 1/x^6 + 2/(x^6(x^2+1)). The second part can be solved by substituting 1/x^2 = t or by partial fractions.
The integrand is \frac{x^2+3}{x^6(x^2+1)} = \frac{(x^2+1)+2}{x^6(x^2+1)} = \frac{1}{x^6} + \frac{2}{x^6(x^2+1)}. Now, \frac{1}{x^6(x^2+1)} = \frac{1}{x^6} \cdot \frac{x^2+1-x^2}{x^2+1} = \frac{1}{x^6} - \frac{1}{x^4(x^2+1)} = \frac{1}{x^6} - \frac{1}{x^4} + \frac{1}{x^2(x^2+1)} = \frac{1}{x^6} - \frac{1}{x^4} + \frac{1}{x^2} - \frac{1}{x^2+1}. Thus, the integral is \int x^{-6} dx + 2 \int (x^{-6} - x^{-4} + x^{-2} - \frac{1}{x^2+1}) dx = \frac{x^{-5}}{-5} + 2(\frac{x^{-5}}{-5} - \frac{x^{-3}}{-3} + \frac{x^{-1}}{-1} - \tan^{-1} x) + C = -\frac{1}{5x^5} - \frac{2}{5x^5} + \frac{2}{3x^3} - \frac{2}{x} - 2\tan^{-1} x + C = -\frac{3}{5x^5} + \frac{2}{3x^3} - \frac{2}{x} - 2\tan^{-1} x + C.
Correct Answer: A