1. Problem statement. We are asked to evaluate the integrals from the quiz. (a) $\int 2x^2 + \frac{2}{x^4} \, dx$. (b) $\int_1^2 (x-2)(x+5) \, dx$. (c) $\int_0^1 (2x+3)\cdot(2x^2+6x-3)^2 \, dx$. (d) $\int \frac{2e^x}{e^x+1} \, dx$. 2. Formulas and important rules. Use the power rule $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ for $n \neq -1$. Use linearity $\int (f+g) \, dx = \int f \, dx + \int g \, dx$. For definite integrals use the Fundamental Theorem $\int_a^b f(x) \, dx = F(b)-F(a)$ where $F' = f$. For substitution use $u$-substitution with $du = u'(x) \, dx$. 3. Solution for 1(a). Rewrite the integrand as $2x^2 + \frac{2}{x^4} = 2x^2 + 2x^{-4}$. Integrate termwise by the power rule. $$\int (2x^2 + 2x^{-4}) \, dx = 2\int x^2 \, dx + 2\int x^{-4} \, dx$$ Compute the antiderivatives. $$2\cdot \frac{x^3}{3} + 2\cdot \frac{x^{-3}}{-3} + C = \frac{2}{3}x^3 - \frac{2}{3}x^{-3} + C$$. 4. Solution for 1(b). Expand the integrand: $(x-2)(x+5) = x^2 +3x -10$. Integrate termwise and form the antiderivative $F(x) = \frac{x^3}{3} + \frac{3x^2}{2} -10x$. Evaluate from 1 to 2 using $F(2)-F(1)$. Compute $F(2) = \frac{8}{3} + 6 -20 = \frac{8}{3} -14 = -\frac{34}{3}$. Compute $F(1) = \frac{1}{3} + \frac{3}{2} -10 = -\frac{49}{6}$. Thus the definite integral is $-\frac{34}{3} - (-\frac{49}{6}) = -\frac{19}{6}$. 5. Solution for 2. Use substitution $u = 2x^2 + 6x -3$ so $du = (4x+6) \, dx = 2(2x+3) \, dx$. Therefore $(2x+3) \, dx = \frac{1}{2} \, du$. The integral becomes $\frac{1}{2}\int_{u(0)}^{u(1)} u^2 \, du$. Compute the limits $u(0) = -3$ and $u(1) = 5$. Integrate to get $\frac{1}{2} \cdot \frac{u^3}{3} \Big|_{-3}^{5} = \frac{1}{6}(5^3 - (-3)^3)$. Simplify $\frac{1}{6}(125 +27) = \frac{152}{6} = \frac{76}{3}$. 6. Solution for 3. Let $u = e^x + 1$ so $du = e^x \, dx$. Then $2e^x \, dx = 2 \, du$ and the integral becomes $\int \frac{2 \, du}{u}$. This integrates to $2\ln|u| + C$ and since $u = e^x+1>0$ we have $2\ln(e^x+1) + C$. 7. Final answers summary. 1(a) $\frac{2}{3}x^3 - \frac{2}{3}x^{-3} + C$. 1(b) $-\frac{19}{6}$. 2 $\frac{76}{3}$. 3 $2\ln(e^x+1) + C$.