Integral Calculus-1
Integral Calculus-1
Allen Star Batch
Grade 12
Question:
Anti-derivative of $$\frac{x - 1}{(x + 1)\sqrt{x^3 + x^2 + x}}$$ is:
$$\tan^{-1}\left(x + \frac{1}{x}\right)$$
$$\tan^{-1}\left(x + \frac{1}{x} + 1\right)$$
$$2\tan^{-1}\left(x + \frac{1}{x}\right)$$
$$\sqrt{x + \frac{1}{x} + 1}$$
Step-by-Step Solution
Key Concept: Recognize that the denominator $\sqrt{x^3 + x^2 + x} = \sqrt{x}\sqrt{x^2 + x + 1}$ and use the substitution $t^2 = x + \frac{1}{x} + 1$ to transform the integral into the standard form $\int \frac{dt}{t^2 + 1} = \tan^{-1}(t)$, accounting for the differential relation $2tdt = (1 - \frac{1}{x^2})dx$.
Rewrite the numerator as $x - 1 = (x+1)^2 - (x^2 + x + 1)$ to split the integral. Substitute $t^2 = x + \frac{1}{x} + 1$, which gives $2tdt = (1 - \frac{1}{x^2})dx$. The integral simplifies to $2\int \frac{tdt}{(t^2+1)t} = 2\int \frac{dt}{t^2+1}$, yielding $2\tan^{-1}(x + \frac{1}{x} + 1) + C$.
Correct Answer: 3