Integral Calculus-1
Integral Calculus-1
Allen Star Batch
Grade 12
Question:
$$\int \frac{x^4 - 2}{x^2\sqrt{x^4 + x^2 + 2}} dx =$$
$$\sqrt{x^2 + 1 + \frac{1}{x^2}} + C$$
$$\sqrt{x^2 + 1 + \frac{2}{x^2}} + C$$
$$\sqrt{x^2 + \frac{1}{x^2}} + C$$
$$\sqrt{x^2 + \frac{2}{x^2}} + C$$
Step-by-Step Solution
Key Concept: Recognizing that the expression under the square root is a perfect square $(x - \frac{1}{x})^2$ allows direct simplification.
Substitute $x^2 + \frac{2}{x^2} + 1 = t^2$ where $t = x - \frac{1}{x}$, giving $\frac{dt}{2} = (x + \frac{1}{x})dx$. This transforms the integral to $\int \frac{dt}{2\sqrt{t}} = \sqrt{t} + C = \sqrt{x^2 + \frac{2}{x^2} + 1} + C$.
Correct Answer: 2