1. **Evaluate the indefinite integral** \( \int \frac{1}{(x+1)^2} \, dx \). Use substitution: let \( u = x + 1 \), so \( du = dx \). Integral becomes \( \int u^{-2} \, du = \int u^{-2} \, du \). Integration rule for \( u^n \) is \( \frac{u^{n+1}}{n+1} \) if \( n \neq -1 \). Here, \( n = -2 \), so: $$\int u^{-2} \, du = \frac{u^{-1}}{-1} + C = -\frac{1}{u} + C = -\frac{1}{x+1} + C.$$ 2. **Find values of \( x \) such that \( f(x) = g(x) \), where \( f(x) = 3x \) and \( g(x) = x^2 + 2 \).** Set equation: $$3x = x^2 + 2$$ Rearranged: $$x^2 - 3x + 2 = 0$$ Factor: $$(x-1)(x-2) = 0$$ Solutions: $$x=1 \text{ or } x=2.$$ 3. **Evaluate limit:** \( \lim_{x \to 0} \frac{x^4}{(1 - \cos 2x)^2} \). Use Taylor expansion near 0 for cosine: $$\cos 2x \approx 1 - 2x^2 + \frac{(2x)^4}{24} = 1 - 2x^2 + \frac{16x^4}{24} = 1 - 2x^2 + \frac{2}{3}x^4.$$ Thus: $$1 - \cos 2x \approx 2x^2 - \frac{2}{3}x^4.$$ Square: $$(1 - \cos 2x)^2 \approx \left(2x^2 - \frac{2}{3}x^4\right)^2 = 4x^4 - \frac{8}{3}x^6 + \frac{4}{9}x^8.$$ Dominant term as \( x \to 0 \) is \( 4x^4 \). Therefore: $$ \lim_{x \to 0} \frac{x^4}{(1 - \cos 2x)^2} = \lim_{x \to 0} \frac{x^4}{4x^4} = \frac{1}{4}. $$ 4. **Find equation of the tangent to the curve \( \ln y + x^3 y = x \) at point \( (1,1) \).** Implicit differentiation: Differentiate both sides w.r.t \( x \): $$\frac{1}{y} \frac{dy}{dx} + 3x^2 y + x^3 \frac{dy}{dx} = 1.$$ Group \( \frac{dy}{dx} \) terms: $$\frac{dy}{dx} \left( \frac{1}{y} + x^3 \right) = 1 - 3x^2 y.$$ Solve for \( \frac{dy}{dx} \): $$\frac{dy}{dx} = \frac{1 - 3x^2 y}{\frac{1}{y} + x^3}.$$ At \( (1,1) \): $$\frac{dy}{dx} = \frac{1 - 3(1)^2(1)}{1 + 1} = \frac{1 - 3}{2} = \frac{-2}{2} = -1.$$ Tangent line equation: $$ y - 1 = -1(x - 1) \Rightarrow y = -x + 2. $$ 5. **Solve inequality:** $$ \frac{3x}{2 + x^2} \leq \frac{1}{x - 1} $$ Bring all to one side: $$ \frac{3x}{2 + x^2} - \frac{1}{x - 1} \leq 0.$$ Common denominator \( (2 + x^2)(x - 1) \): $$ \frac{3x(x-1) - (2 + x^2)}{(2 + x^2)(x - 1)} \leq 0.$$ Simplify numerator: $$3x^2 - 3x - 2 - x^2 = 2x^2 - 3x - 2.$$ Factor numerator: $$2x^2 - 3x - 2 = (2x + 1)(x - 2).$$ Inequality: $$ \frac{(2x + 1)(x - 2)}{(2 + x^2)(x - 1)} \leq 0.$$ Denominator \( (2 + x^2) > 0 \) always. Critical points: numerator zeros at \( x = -\frac{1}{2}, 2 \); denominator zero at \( x = 1 \) (excluded). Test intervals: - \( (-\infty, -\frac{1}{2}) \): numerator positive, denominator negative (because \(x-1<0\)), ratio negative. - \( (-\frac{1}{2}, 1) \): numerator negative, denominator negative, ratio positive. - \( (1, 2) \): numerator negative, denominator positive, ratio negative. - \( (2, \infty) \): numerator positive, denominator positive, ratio positive. Include points where numerator zero (inequality \( \leq 0 \)) Excluded \( x = 1 \) (denominator zero). Solution intervals: $$ \left(-\infty, -\frac{1}{2} \right] \cup (1, 2]. $$ 6. **Prove sum of two odd integers is even and product is odd.** Let two odd integers be \(2a + 1\) and \(2b + 1\) where \(a,b \in \mathbb{Z}\). Sum: $$ (2a + 1) + (2b + 1) = 2a + 2b + 2 = 2(a + b + 1), $$ which is divisible by 2, so sum is even. Product: $$ (2a + 1)(2b + 1) = 4ab + 2a + 2b + 1 = 2(2ab + a + b) + 1, $$ odd by form \(2k+1\). Hence product is odd. 7. **Find critical points of \( f(x) = x^3 - 12x -16 \) and classify them.** Compute derivative: $$ f'(x) = 3x^2 - 12. $$ Critical points where \( f'(x) = 0 \): $$ 3x^2 - 12 = 0 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2. $$ Second derivative: $$ f''(x) = 6x. $$ At \( x = -2 \): \( f''(-2) = -12 < 0 \), so local maximum. At \( x = 2 \): \( f''(2) = 12 > 0 \), so local minimum. 8. **Evaluate** $$ \int_0^1 \sin^{-1}\left( \frac{2x}{1+x^2} \right) dx.$$ Recognize: $$ \frac{2x}{1+x^2} = \sin(2 \tan^{-1} x), $$ because \( \sin 2\theta = 2 \sin \theta \cos \theta \), and \( \sin \theta = \frac{x}{\sqrt{1+x^2}} \), \( \cos \theta = \frac{1}{\sqrt{1+x^2}} \), so: $$ \sin (2 \tan^{-1} x) = \frac{2x}{1+x^2}. $$ Thus inside integrand is: $$ \sin^{-1}(\sin (2 \tan^{-1} x)) = 2 \tan^{-1} x, $$ for \( x \in [0,1] \). Integral becomes: $$ \int_0^1 2 \tan^{-1} x \, dx = 2 \int_0^1 \tan^{-1} x \, dx.$$ Use integration by parts: Let \( u = \tan^{-1} x \), \( dv = dx \), so \( du = \frac{1}{1+x^2} dx \), \( v = x \). $$ \int \tan^{-1} x \, dx = x \tan^{-1} x - \int \frac{x}{1+x^2} dx.$$ Evaluate \( \int \frac{x}{1+x^2} dx \): let \( w = 1 + x^2 \), \( dw = 2x dx \), then $$ \int \frac{x}{1+x^2} dx = \frac{1}{2} \int \frac{dw}{w} = \frac{1}{2} \ln |1+x^2| + C.$$ So: $$ \int \tan^{-1} x \, dx = x \tan^{-1} x - \frac{1}{2} \ln (1+x^2) + C.$$ Evaluate definite integral: $$ \int_0^1 \tan^{-1} x \, dx = \left[x \tan^{-1} x - \frac{1}{2} \ln (1+x^2) \right]_0^1. $$ At \( x=1 \): \( 1 \times \frac{\pi}{4} - \frac{1}{2} \ln 2 = \frac{\pi}{4} - \frac{1}{2} \ln 2. \) At \( x=0 \): 0. Therefore: $$ \int_0^1 2 \tan^{-1} x \, dx = 2 \left( \frac{\pi}{4} - \frac{1}{2} \ln 2 \right) = \frac{\pi}{2} - \ln 2. $$ 9. **Using Mean-Value Theorem, prove:** $$\frac{b-a}{\sqrt{(1-a)(1+a)}} < \sin^{-1}(b) - \sin^{-1}(a) < \frac{b-a}{\sqrt{(1-b)(1+b)}}, \quad 0 < a < b < 1.$$ Let \( f(x) = \sin^{-1}(x) \), continuous and differentiable on \( [a,b] \). By MVT, $$ f'(c) = \frac{f(b)-f(a)}{b-a} $$ for some \( c \in (a,b) \). Derivative: $$ f'(x) = \frac{1}{\sqrt{1 - x^2}} = \frac{1}{\sqrt{(1-x)(1+x)}}. $$ Since \( f'(x) \) is increasing on \( (0,1) \), $$ f'(a) < f'(c) < f'(b), $$ implying $$ \frac{1}{\sqrt{(1-a)(1+a)}} < \frac{f(b) - f(a)}{b - a} < \frac{1}{\sqrt{(1-b)(1+b)}}. $$ Multiply both sides by \( b-a > 0 \): $$ \frac{b-a}{\sqrt{(1-a)(1+a)}} < \sin^{-1}(b) - \sin^{-1}(a) < \frac{b-a}{\sqrt{(1-b)(1+b)}}. $$ 10. **Show:** $$ \frac{\pi}{6} + \frac{1}{5\sqrt{3}} < \sin^{-1}\left(\frac{3}{5}\right) < \frac{1}{8} + \frac{\pi}{6}. $$ Set \( a = \frac{1}{2} \), \( b = \frac{3}{5} \). Use inequality: Lower bound: $$ \sin^{-1}(b) > \sin^{-1}(a) + \frac{b-a}{\sqrt{(1-a)(1+a)}}. $$ Calculate: $$ \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}, \quad b-a = \frac{3}{5} - \frac{1}{2} = \frac{6-5}{10} = \frac{1}{10}. $$ Compute denominator: $$ \sqrt{\left(1-\frac{1}{2}\right) \left(1+\frac{1}{2}\right)} = \sqrt{\frac{1}{2} \times \frac{3}{2}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}. $$ Therefore: $$ \sin^{-1}\left(\frac{3}{5}\right) > \frac{\pi}{6} + \frac{\frac{1}{10}}{\frac{\sqrt{3}}{2}} = \frac{\pi}{6} + \frac{1}{10} \times \frac{2}{\sqrt{3}} = \frac{\pi}{6} + \frac{1}{5\sqrt{3}}. $$ Upper bound: $$ \sin^{-1}(b) < \sin^{-1}(a) + \frac{b-a}{\sqrt{(1-b)(1+b)}}. $$ Compute denominator: $$ \sqrt{\left(1-\frac{3}{5}\right) \left(1+\frac{3}{5}\right)} = \sqrt{\frac{2}{5} \times \frac{8}{5}} = \sqrt{\frac{16}{25}} = \frac{4}{5}. $$ Thus: $$ \sin^{-1}\left(\frac{3}{5}\right) < \frac{\pi}{6} + \frac{\frac{1}{10}}{\frac{4}{5}} = \frac{\pi}{6} + \frac{1}{10} \times \frac{5}{4} = \frac{\pi}{6} + \frac{1}{8}. $$ Hence desired inequalities are proved. Final answers: (a) \( - \frac{1}{x+1} + C \) (b) \( x=1,2 \) (c) \( \frac{1}{4} \) (d) \( y = -x + 2 \) (e) \( (-\infty, -\frac{1}{2}] \cup (1,2] \) (f) Sum even, product odd proved. (g) Critical points at \( x = -2 \) (local max), \( x = 2 \) (local min). (h) \( \frac{\pi}{2} - \ln 2 \) (i,j) Mean Value Theorem inequalities and numerical bounds showed as above.