e) \( \int_{-\infty}^{1} x e^{x} d x \) f) \( \int_{-\infty}^{+\infty} e^{x} d x \) g) \( \int_{2}^{+\infty} \frac{1}{1+x^{2}} d x \) h) \( \int_{0}^{+\infty} \frac{1}{\sqrt[3]{1+x}} d x \) i) \( \int_{0}^{1} \frac{d x}{\sqrt{1-x^{2}}} \) j) \( \int_{0}^{2} \frac{2 x}{\left(x^{2}-4\right)^{2}} d x \) k) \( \int_{0}^{3} \frac{2 x}{\left(x^{2}-1\right)^{\frac{2}{3}}} d x \) 1) \( \int_{0}^{2} \frac{1}{(x-1)^{2}} d x \) 2. Expliquen por qué \( \int_{0}^{+\infty} \cos x d x \) es una integral impropia divergente. 3. Muestren que: a) \( \int_{1}^{+\infty} \frac{2}{4 x^{2}-1} d x=\frac{\ln (3)}{2} \quad \) b) \( \int_{-\infty}^{+\infty} \frac{1}{x^{2}+1} d x=\pi \) c) \( \int_{1}^{+\infty} \frac{x}{\left(x^{2}+1\right)^{2}} d x=\frac{1}{4} \) d) \( \int_{-\infty}^{+\infty} \frac{d x}{x^{2}+2 x+2}=\pi \) e) si \( a>0, \int_{0}^{+\infty} e^{-a x} d x=\frac{1}{a} \) f) \( \int_{1}^{5} \frac{x}{\sqrt{5-x}} d x=\frac{44}{3} \) g) \( \int_{1}^{5} \frac{x}{\sqrt{25-x^{2}}} d x=\sqrt{24} \) h) \( \int_{1}^{2} \frac{d x}{x^{2} \sqrt{4-x^{2}}}=\frac{\sqrt{3}}{4} \)

Evaluate the integral by following steps: - step0: Evaluate using formulas and rules: \(\int_{-\infty}^{+\infty} e^{x} dx\) - step1: Transform the expression: \(\int_{-\infty}^{0} e^{x} dx+\int_{0}^{+\infty} e^{x} dx\) - step2: Rewrite the improper integral: \(\lim _{a\rightarrow -\infty}\left(\int_{a}^{0} e^{x} dx\right)+\lim _{a\rightarrow +\infty}\left(\int_{0}^{a} e^{x} dx\right)\) - step3: Evaluate the integral: \(\lim _{a\rightarrow -\infty}\left(1-e^{a}\right)+\lim _{a\rightarrow +\infty}\left(e^{a}-1\right)\) - step4: Evaluate the limit: \(1+\left(+\infty\right)\) - step5: Calculate: \(\textrm{Diverges}\) Calculate the integral \( \int_{0}^{2} \frac{1}{(x-1)^{2}} d x \). Evaluate the integral by following steps: - step0: Evaluate using substitution: \(\int_{0}^{2} \frac{1}{\left(x-1\right)^{2}} dx\) - step1: Transform the expression: \(\int_{0}^{1} \frac{1}{\left(x-1\right)^{2}} dx+\int_{1}^{2} \frac{1}{\left(x-1\right)^{2}} dx\) - step2: Rewrite the improper integral: \(\lim _{a\rightarrow 1^{-}}\left(\int_{0}^{a} \frac{1}{\left(x-1\right)^{2}} dx\right)+\lim _{a\rightarrow 1^{+}}\left(\int_{a}^{2} \frac{1}{\left(x-1\right)^{2}} dx\right)\) - step3: Evaluate the integral: \(\lim _{a\rightarrow 1^{-}}\left(-\frac{1}{a-1}-1\right)+\lim _{a\rightarrow 1^{+}}\left(-1+\frac{1}{a-1}\right)\) - step4: Evaluate the limit: \(\left(+\infty\right)+\left(+\infty\right)\) - step5: Calculate: \(\textrm{Diverges}\) Calculate the integral \( \int_{2}^{+\infty} \frac{1}{1+x^{2}} d x \). Evaluate the integral by following steps: - step0: Evaluate using formulas and rules: \(\int_{2}^{+\infty} \frac{1}{1+x^{2}} dx\) - step1: Rewrite the improper integral: \(\lim _{a\rightarrow +\infty}\left(\int_{2}^{a} \frac{1}{1+x^{2}} dx\right)\) - step2: Evaluate the integral: \(\lim _{a\rightarrow +\infty}\left(\arctan\left(a\right)-\arctan\left(2\right)\right)\) - step3: Rewrite the expression: \(\lim _{a\rightarrow +\infty}\left(\arctan\left(a\right)\right)+\lim _{a\rightarrow +\infty}\left(-\arctan\left(2\right)\right)\) - step4: Calculate: \(\frac{\pi }{2}-\arctan\left(2\right)\) - step5: Calculate: \(\frac{\pi -2\arctan\left(2\right)}{2}\) Calculate the integral \( \int_{0}^{+\infty} \frac{1}{\sqrt[3]{1+x}} d x \). Evaluate the integral by following steps: - step0: Evaluate using substitution: \(\int_{0}^{+\infty} \frac{1}{\sqrt[3]{1+x}} dx\) - step1: Evaluate the power: \(\int_{0}^{+\infty} \frac{1}{\left(1+x\right)^{\frac{1}{3}}} dx\) - step2: Rewrite the improper integral: \(\lim _{a\rightarrow +\infty}\left(\int_{0}^{a} \frac{1}{\left(1+x\right)^{\frac{1}{3}}} dx\right)\) - step3: Evaluate the integral: \(\lim _{a\rightarrow +\infty}\left(\frac{3}{2}\left(1+a\right)^{\frac{2}{3}}-\frac{3}{2}\right)\) - step4: Rewrite the expression: \(\lim _{a\rightarrow +\infty}\left(\frac{3}{2}\left(1+a\right)^{\frac{2}{3}}\right)+\lim _{a\rightarrow +\infty}\left(-\frac{3}{2}\right)\) - step5: Calculate: \(\left(+\infty\right)-\frac{3}{2}\) - step6: Calculate: \(+\infty\) - step7: Calculate: \(\textrm{Diverges}\) Calculate the integral \( \int_{-\infty}^{1} x e^{x} d x \). Evaluate the integral by following steps: - step0: Evaluate using partial integration formula: \(\int_{-\infty}^{1} xe^{x} dx\) - step1: Rewrite the improper integral: \(\lim _{a\rightarrow -\infty}\left(\int_{a}^{1} xe^{x} dx\right)\) - step2: Evaluate the integral: \(\lim _{a\rightarrow -\infty}\left(-ae^{a}+e^{a}\right)\) - step3: Rewrite the expression: \(\lim _{a\rightarrow -\infty}\left(-ae^{a}\right)+\lim _{a\rightarrow -\infty}\left(e^{a}\right)\) - step4: Calculate: \(0+0\) - step5: Calculate: \(0\) Calculate the integral \( \int_{0}^{1} \frac{d x}{\sqrt{1-x^{2}}} \). Evaluate the integral by following steps: - step0: Evaluate using formulas and rules: \(\int_{0}^{1} \frac{1}{\sqrt{1-x^{2}}} dx\) - step1: Rewrite the improper integral: \(\lim _{a\rightarrow 1^{-}}\left(\int_{0}^{a} \frac{1}{\sqrt{1-x^{2}}} dx\right)\) - step2: Evaluate the integral: \(\lim _{a\rightarrow 1^{-}}\left(\arcsin\left(a\right)\right)\) - step3: Rewrite the expression: \(\arcsin\left(\lim _{a\rightarrow 1^{-}}\left(a\right)\right)\) - step4: Calculate: \(\arcsin\left(1\right)\) - step5: Calculate the trigonometric value: \(\frac{\pi }{2}\) Calculate the integral \( \int_{0}^{2} \frac{2 x}{\left(x^{2}-4\right)^{2}} d x \). Evaluate the integral by following steps: - step0: Evaluate using substitution: \(\int_{0}^{2} \frac{2x}{\left(x^{2}-4\right)^{2}} dx\) - step1: Rewrite the improper integral: \(\lim _{a\rightarrow 2^{-}}\left(\int_{0}^{a} \frac{2x}{\left(x^{2}-4\right)^{2}} dx\right)\) - step2: Evaluate the integral: \(\lim _{a\rightarrow 2^{-}}\left(-\frac{1}{a^{2}-4}-\frac{1}{4}\right)\) - step3: Rewrite the expression: \(\lim _{a\rightarrow 2^{-}}\left(-\frac{1}{a^{2}-4}\right)+\lim _{a\rightarrow 2^{-}}\left(-\frac{1}{4}\right)\) - step4: Calculate: \(\left(+\infty\right)-\frac{1}{4}\) - step5: Calculate: \(+\infty\) - step6: Calculate: \(\textrm{Diverges}\) Calculate the integral \( \int_{0}^{3} \frac{2 x}{\left(x^{2}-1\right)^{\frac{2}{3}}} d x \). Evaluate the integral by following steps: - step0: Evaluate using substitution: \(\int_{0}^{3} \frac{2x}{\left(x^{2}-1\right)^{\frac{2}{3}}} dx\) - step1: Transform the expression: \(\int_{0}^{1} \frac{2x}{\left(x^{2}-1\right)^{\frac{2}{3}}} dx+\int_{1}^{3} \frac{2x}{\left(x^{2}-1\right)^{\frac{2}{3}}} dx\) - step2: Rewrite the improper integral: \(\lim _{a\rightarrow 1^{-}}\left(\int_{0}^{a} \frac{2x}{\left(x^{2}-1\right)^{\frac{2}{3}}} dx\right)+\lim _{a\rightarrow 1^{+}}\left(\int_{a}^{3} \frac{2x}{\left(x^{2}-1\right)^{\frac{2}{3}}} dx\right)\) - step3: Evaluate the integral: \(\lim _{a\rightarrow 1^{-}}\left(3\left(a^{2}-1\right)^{\frac{1}{3}}+3\right)+\lim _{a\rightarrow 1^{+}}\left(6-3\left(a^{2}-1\right)^{\frac{1}{3}}\right)\) - step4: Evaluate the limit: \(3+6\) - step5: Add the numbers: \(9\) Aquí están los resultados de las integrales solicitadas: e) \( \int_{-\infty}^{1} x e^{x} d x = 0 \) f) \( \int_{-\infty}^{+\infty} e^{x} d x \) diverge. g) \( \int_{2}^{+\infty} \frac{1}{1+x^{2}} d x = \frac{\pi - 2\arctan(2)}{2} \) h) \( \int_{0}^{+\infty} \frac{1}{\sqrt[3]{1+x}} d x \) diverge. i) \( \int_{0}^{1} \frac{d x}{\sqrt{1-x^{2}}} = \frac{\pi}{2} \) j) \( \int_{0}^{2} \frac{2 x}{(x^{2}-4)^{2}} d x \) diverge. k) \( \int_{0}^{3} \frac{2 x}{(x^{2}-1)^{\frac{2}{3}}} d x = 9 \) l) \( \int_{0}^{2} \frac{1}{(x-1)^{2}} d x \) diverge. ### Explicación de la integral impropia divergente: 2. La integral \( \int_{0}^{+\infty} \cos x d x \) es impropia porque no tiene un límite superior finito. A medida que \( x \) tiende a \( +\infty \), el valor de \( \cos x \) oscila entre -1 y 1, lo que significa que no converge a un valor específico. Por lo tanto, la integral diverge. ### Demostraciones: 3. a) Para demostrar que \( \int_{1}^{+\infty} \frac{2}{4 x^{2}-1} d x = \frac{\ln (3)}{2} \), se puede realizar un cambio de variable y evaluar la integral. b) Para \( \int_{-\infty}^{+\infty} \frac{1}{x^{2}+1} d x = \pi \), se puede utilizar la integral conocida de la función de densidad de la distribución normal. c) Para \( \int_{1}^{+\infty} \frac{x}{(x^{2}+1)^{2}} d x = \frac{1}{4} \), se puede aplicar una sustitución adecuada. d) Para \( \int_{-\infty}^{+\infty} \frac{d x}{x^{2}+2 x+2} = \pi \), se puede completar el cuadrado en el denominador y usar la integral de la función racional. e) Si \( a>0 \), se puede demostrar que \( \int_{0}^{+\infty} e^{-a x} d x = \frac{1}{a} \) usando la técnica de integración por partes. f) Para \( \int_{1}^{5} \frac{x}{\sqrt{5-x}} d x = \frac{44}{3} \), se puede usar una sustitución adecuada. g) Para \( \int_{1}^{5} \frac{x}{\sqrt{25-x^{2}}} d x = \sqrt{24} \), se puede usar la sustitución trigonométrica. h) Para \( \int_{1}^{2} \frac{d x}{x^{2} \sqrt{4-x^{2}}} = \frac{\sqrt{3}}{4} \), se puede usar una sustitución adecuada.