If function $y = f(x)$ is continuous at $x = c$ such that $f(c) \neq 0$, then $f(x)f(c) > 0\forall x \in (c-h, c+h)$ where $h$ is sufficiently small positive quantity
$\lim_{n\to\infty} \ln\left[\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)...\left(1+\frac{n}{n}\right)\right] = 1 + 2\ln 2$
Let $f$ be a continuous and non-negative function defined on $[a, b]$. If $\int_a^b f(x)dx = 0$, then $f(x) = 0$ $\forall x \in [a,b]$
Let $f$ be a continuous function defined on $[a, b]$. If $\int_a^b f(x)dx = 0$, then there exists at least one $c \in (a,b)$ for which $f(c) = 0$
Step-by-Step Solution
Key Concept: Decompose inverse trigonometric expressions into recognizable forms and use partial fractions with the factorization $1-x^8=(1-x^4)(1+x^4)$.
The integral $\int \frac{1}{1-x^8}\left[\cos^{-1}\left(\frac{2x}{1+x^2}\right) + \tan^{-1}\left(\frac{2x}{1-x^2}\right)\right]dx$ is evaluated by recognizing that $\cos^{-1}\left(\frac{2x}{1+x^2}\right) = \frac{\pi}{2} - \sin^{-1}\left(\frac{2x}{1+x^2}\right)$ and $\tan^{-1}\left(\frac{2x}{1-x^2}\right)$ relates to $2\tan^{-1}(x)$. After simplification using partial fractions and trigonometric identities, the result involves logarithmic and inverse tangent terms combined with radicals.
Correct Answer: 1,2