Inmediatas:
$$\int x^5 \, dx = \frac{x^6}{6} + C$$
$$\int e^{2x} \, dx = \frac{1}{2}e^{2x} + C$$
$$\int x^{-3} \, dx = \frac{x^{-2}}{-2} + C = -\frac{1}{2x^2} + C$$
$$\int \sqrt{x} \, dx = \int x^{1/2} \, dx = \frac{x^{3/2}}{3/2} + C = \frac{2}{3}\sqrt{x^3} + C$$
$$\int \frac{1}{\sqrt[3]{x}} \, dx = \int x^{-1/3} \, dx = \frac{x^{2/3}}{2/3} + C = \frac{3}{2}\sqrt[3]{x^2} + C$$
$$\int 7x^2 \, dx = 7 \int x^2 \, dx = \frac{7x^3}{3} + C$$
$$\int 5^x \, dx = \frac{5^x}{\ln 5} + C$$
$$\int \frac{1}{x} \, dx = \ln|x| + C$$
$$\int \frac{3}{x} \, dx = 3 \ln|x| + C$$
$$\int e^{x+2} \, dx = e^{x+2} + C$$
$$\int \frac{1}{x} \, dx = \ln|x| + C$$
$$\int (x^2 + x + 1) \, dx = \frac{x^3}{3} + \frac{x^2}{2} + x + C$$
$$\int \frac{x+1}{x} \, dx = \int (1 + \frac{1}{x}) \, dx = x + \ln|x| + C$$
$$\int (x-2)^2 \, dx = \int (x^2 – 4x + 4) \, dx = \int x^2 \, dx – 4 \int x \, dx + 4 \int dx = \frac{x^3}{3} – 4 \frac{x^2}{2} + 4x + C = \frac{x^3}{3} – 2x^2 + 4x + C$$
$$\int \frac{1}{x^4} \, dx = \int x^{-4} \, dx = \frac{x^{-3}}{-3} + C = -\frac{1}{3} \frac{1}{x^3} + C$$
$$\int \frac{x^2 – x + 5}{x} \, dx = \int \left( x – 1 + \frac{5}{x} \right) \, dx = \int x \, dx – \int dx + 5 \int \frac{1}{x} \, dx = \frac{x^2}{2} – x + 5 \ln|x| + C$$
$$\int \sqrt[3]{x^2} \, dx = \int x^{2/3} \, dx = \frac{x^{5/3}}{5/3} + C = \frac{3}{5} x^{5/3} + C = \frac{3}{5} \sqrt[3]{x^5} + C = \frac{3}{5} x \sqrt[3]{x^2} + C$$
$$\int \frac{1}{5x – 2} \, dx = \frac{1}{5} \int \frac{5}{5x – 2} \, dx = \frac{1}{5} \ln|5x – 2| + C$$
$$\int e^{-3x} \, dx = -\frac{1}{3} \int -3e^{-3x} \, dx = -\frac{1}{3}e^{-3x} + C$$
$$\int (2x + 7)^4 \, dx = \frac{1}{2} \int 2(2x + 7)^4 \, dx = \frac{1}{2} \frac{(2x + 7)^5}{5} + C = \frac{(2x + 7)^5}{10} + C$$
$$\int \sqrt{4x – 1} \, dx = \frac{1}{4} \int 4(4x – 1)^{1/2} \, dx = \frac{1}{4} \frac{(4x – 1)^{3/2}}{3/2} + C = \frac{1}{6} \sqrt{(4x – 1)^3} + C$$
$$\int 3^{2x} \, dx = \frac{1}{2} \int 2 \cdot 3^{2x} \, dx = \frac{1}{2} \frac{3^{2x}}{\ln 3} + C$$
$$\int \frac{1}{\sqrt{5x + 2}} \, dx = \frac{1}{5} \int 5(5x + 2)^{-1/2} \, dx = \frac{1}{5} \frac{(5x + 2)^{1/2}}{1/2} + C = \frac{2}{5} \sqrt{5x + 2} + C$$
$$\int (e^x + e^{-x})^2 \, dx = \int (e^{2x} + 2 + e^{-2x}) \, dx = \frac{1}{2}e^{2x} + 2x – \frac{1}{2}e^{-2x} + C$$
$$\int \frac{1}{x \sqrt[4]{x}} \, dx = \int x^{-5/4} \, dx = \frac{x^{-1/4}}{-1/4} + C = -\frac{4}{\sqrt[4]{x}} + C$$
$$\int \frac{dx}{7 – x} = -\int \frac{-1}{7 – x} \, dx = -\ln|7 – x| + C$$
$$\int \frac{x + 3}{\sqrt{x}} \, dx = \int (x^{1/2} + 3x^{-1/2}) \, dx = \frac{2}{3}x^{3/2} + 6x^{1/2} + C$$
$$\int \frac{x^3 – 1}{x^2} \, dx = \int (x – x^{-2}) \, dx = \frac{x^2}{2} + \frac{1}{x} + C$$
$$\int e^{2x+3} \, dx = \frac{1}{2} \int 2e^{2x+3} \, dx = \frac{1}{2}e^{2x+3} + C$$
$$\int \frac{1}{1 – 3x} \, dx = -\frac{1}{3} \int \frac{-3}{1 – 3x} \, dx = -\frac{1}{3}\ln|1 – 3x| + C$$
$$\int \frac{1}{(4x + 1)^3} \, dx = \frac{1}{4} \int 4(4x + 1)^{-3} \, dx = \frac{(4x + 1)^{-2}}{4(-2)} + C = -\frac{1}{8(4x + 1)^2} + C$$
$$\int \sqrt[3]{5x – 2} \, dx = \frac{1}{5} \int 5(5x – 2)^{1/3} \, dx = \frac{1}{5} \frac{(5x – 2)^{4/3}}{4/3} + C = \frac{3}{20}\sqrt[3]{(5x – 2)^4} + C$$
$$\int \frac{dx}{\sqrt{1 – 2x}} = -\frac{1}{2} \int -2(1 – 2x)^{-1/2} \, dx = -\frac{1}{2} \frac{(1 – 2x)^{1/2}}{1/2} + C = -\sqrt{1 – 2x} + C$$
$$\int (\sqrt{x} – 1)^2 \, dx = \int (x – 2\sqrt{x} + 1) \, dx = \frac{x^2}{2} – \frac{4}{3}x^{3/2} + x + C$$
Por cambio de variable:
$$\int x\sqrt{x-1} \, dx = \begin{pmatrix} t = x-1 \\ dt = dx \end{pmatrix} = \int (t+1)\sqrt{t} \, dt = \int t\sqrt{t} \, dt + \int \sqrt{t} \, dt = \int t \cdot t^{1/2} \, dt + \int t^{1/2} \, dt = \int t^{3/2} \, dt + \int t^{1/2} \, dt = \frac{t^{5/2}}{5/2} + \frac{t^{3/2}}{3/2} + C = \frac{2}{5} t^{5/2} + \frac{2}{3} t^{3/2} + C = \frac{2}{5} (x-1)^{5/2} + \frac{2}{3} (x-1)^{3/2} + C$$
$$\int \frac{x^2}{x^3 – 2} \, dx = \begin{pmatrix} t = x^3 – 2 \\ dt = 3x^2 \, dx \end{pmatrix} = \frac{1}{3} \int \frac{dt}{t} = \frac{1}{3} \ln|t| + C = \frac{1}{3} \ln|x^3 – 2| + C$$
$$\int (e^x – 3)^4 e^x \, dx = \begin{pmatrix} t = e^x – 3 \\ dt = e^x \, dx \end{pmatrix} = \int t^4 \, dt = \frac{t^5}{5} + C = \frac{(e^x – 3)^5}{5} + C$$
$$\int \frac{\ln x}{x} \, dx = \begin{pmatrix} t = \ln x \\ dt = \frac{1}{x} \, dx \end{pmatrix} = \int t \, dt = \frac{t^2}{2} + C = \frac{(\ln |x|)^2}{2} + C$$
$$\int x\sqrt{x^2 + 5} \, dx = \left( \begin{matrix} t = x^2+5 \\ dt = 2x \, dx \end{matrix} \right) = \frac{1}{2} \int \sqrt{t} \, dt = \frac{1}{2} \frac{t^{3/2}}{3/2} + C = \frac{1}{3}(x^2+5)^{3/2} + C$$
$$\int x e^{x^2} \, dx = \left( \begin{matrix} t = x^2 \\ dt = 2x \, dx \end{matrix} \right) = \frac{1}{2} \int e^t \, dt = \frac{1}{2} e^t + C = \frac{1}{2} e^{x^2} + C$$
$$\int \frac{(\ln x)^3}{x} \, dx = \left( \begin{matrix} t = \ln x \\ dt = \frac{1}{x} dx \end{matrix} \right) = \int t^3 \, dt = \frac{t^4}{4} + C = \frac{(\ln x)^4}{4} + C$$
$$\int \frac{x^3}{(x^4 + 1)^2} \, dx = \left( \begin{matrix} t = x^4+1 \\ dt = 4x^3 dx \end{matrix} \right) = \frac{1}{4} \int t^{-2} \, dt = -\frac{1}{4t} + C = -\frac{1}{4(x^4+1)} + C$$
$$\int \frac{e^x}{\sqrt[3]{e^x + 2}} \, dx = \left( \begin{matrix} t = e^x+2 \\ dt = e^x dx \end{matrix} \right) = \int t^{-1/3} \, dt = \frac{t^{2/3}}{2/3} + C = \frac{3}{2}(e^x+2)^{2/3} + C$$
$$\int \frac{1}{x \sqrt{\ln x}} \, dx = \left( \begin{matrix} t = \ln x \\ dt = \frac{1}{x} dx \end{matrix} \right) = \int t^{-1/2} \, dt = 2\sqrt{t} + C = 2\sqrt{\ln x} + C$$
$$\int (x-1)(x^2 – 2x + 5)^4 \, dx = \left( \begin{matrix} t = x^2-2x+5 \\ dt = (2x-2) dx \end{matrix} \right) = \frac{1}{2} \int t^4 \, dt = \frac{t^5}{10} + C = \frac{(x^2-2x+5)^5}{10} + C$$
$$\int \frac{e^{\sqrt{x}}}{\sqrt{x}} \, dx = \left( \begin{matrix} t = \sqrt{x} \\ dt = \frac{1}{2\sqrt{x}} dx \end{matrix} \right) = 2 \int e^t \, dt = 2e^t + C = 2e^{\sqrt{x}} + C$$
$$\int \frac{1}{x (\ln x)^2} \, dx = \left( \begin{matrix} t = \ln x \\ dt = \frac{1}{x} dx \end{matrix} \right) = \int t^{-2} \, dt = -\frac{1}{t} + C = -\frac{1}{\ln x} + C$$
$$\int \frac{x+1}{x^2+2x+3} \, dx = \left( \begin{matrix} t = x^2+2x+3 \\ dt = (2x+2) dx \end{matrix} \right) = \frac{1}{2} \int \frac{1}{t} dt = \frac{1}{2} \ln|t| + C = \frac{1}{2} \ln|x^2+2x+3| + C$$
$$\int \frac{e^x}{e^x + 4} \, dx = \left( \begin{matrix} t = e^x + 4 \\ dt = e^x dx \end{matrix} \right) = \int \frac{1}{t} dt = \ln|t| + C = \ln(e^x + 4) + C$$
$$\int x^2 \sqrt{x^3 – 1} \, dx = \left( \begin{matrix} t = x^3 – 1 \\ dt = 3x^2 dx \end{matrix} \right) = \frac{1}{3} \int \sqrt{t} \, dt = \frac{1}{3} \frac{t^{3/2}}{3/2} + C = \frac{2}{9}(x^3 – 1)^{3/2} + C$$
$$\int \frac{\sqrt{\ln x}}{x} \, dx = \left( \begin{matrix} t = \ln x \\ dt = \frac{1}{x} dx \end{matrix} \right) = \int \sqrt{t} \, dt = \frac{t^{3/2}}{3/2} + C = \frac{2}{3}(\ln x)^{3/2} + C$$
$$\int (3x + 2)^5 \, dx = \left( \begin{matrix} t = 3x + 2 \\ dt = 3 dx \end{matrix} \right) = \frac{1}{3} \int t^5 \, dt = \frac{1}{3} \frac{t^6}{6} + C = \frac{(3x + 2)^6}{18} + C$$
$$\int x^2 e^{x^3 + 1} \, dx = \left( \begin{matrix} t = x^3 + 1 \\ dt = 3x^2 dx \end{matrix} \right) = \frac{1}{3} \int e^t \, dt = \frac{1}{3} e^t + C = \frac{1}{3} e^{x^3 + 1} + C$$
$$\int \frac{1}{x (\ln x)^4} \, dx = \left( \begin{matrix} t = \ln x \\ dt = \frac{1}{x} dx \end{matrix} \right) = \int t^{-4} \, dt = \frac{t^{-3}}{-3} + C = -\frac{1}{3(\ln x)^3} + C$$
$$\int \frac{x}{\sqrt{x^2 – 9}} \, dx = \left( \begin{matrix} t = x^2 – 9 \\ dt = 2x dx \end{matrix} \right) = \frac{1}{2} \int t^{-1/2} \, dt = \frac{1}{2} (2\sqrt{t}) + C = \sqrt{x^2 – 9} + C$$
$$\int e^{-x} (1 – e^{-x})^2 \, dx = \left( \begin{matrix} t = 1 – e^{-x} \\ dt = e^{-x} dx \end{matrix} \right) = \int t^2 \, dt = \frac{t^3}{3} + C = \frac{(1 – e^{-x})^3}{3} + C$$
$$\int \frac{x^4}{x^5 + 7} \, dx = \left( \begin{matrix} t = x^5 + 7 \\ dt = 5x^4 dx \end{matrix} \right) = \frac{1}{5} \int \frac{1}{t} dt = \frac{1}{5} \ln|t| + C = \frac{1}{5} \ln|x^5 + 7| + C$$
$$\int x \sqrt[3]{1 – x^2} \, dx = \left( \begin{matrix} t = 1 – x^2 \\ dt = -2x dx \end{matrix} \right) = -\frac{1}{2} \int t^{1/3} \, dt = -\frac{1}{2} \frac{t^{4/3}}{4/3} + C = -\frac{3}{8}(1 – x^2)^{4/3} + C$$
Ejercicio: Calcula las siguientes integrales por partes:
$$\int \ln x \, dx = \left( \begin{matrix} u = \ln x \Rightarrow du = \frac{dx}{x} \\ dv = dx \Rightarrow v = x \end{matrix} \right) = x \ln x – \int x \frac{dx}{x} = x \ln x – \int dx = x \ln x – x + C$$
$$\int xe^x \, dx = \left( \begin{matrix} u = x \Rightarrow du = dx \\ dv = e^x \, dx \Rightarrow v = e^x \end{matrix} \right) = xe^x – \int e^x \, dx = xe^x – e^x + C$$
$$\int x \ln x \, dx = \left( \begin{matrix} u = \ln x \Rightarrow du = \frac{dx}{x} \\ dv = x \, dx \Rightarrow v = \frac{x^2}{2} \end{matrix} \right) = \frac{x^2}{2} \ln x – \frac{1}{2} \int x^2 \frac{dx}{x} = \frac{x^2}{2} \ln x – \frac{1}{2} \int x \, dx = \frac{x^2}{2} \ln x – \frac{1}{2} \frac{x^2}{2} + C = \frac{x^2}{2} \ln x – \frac{x^2}{4} + C$$
$$\int x^2 e^x \, dx = \left( \begin{matrix} u = x^2 \Rightarrow du = 2x \, dx \\ dv = e^x \, dx \Rightarrow v = e^x \end{matrix} \right) = x^2 e^x – 2 \int xe^x \, dx = \left( \begin{matrix} u = x \Rightarrow du = dx \\ dv = e^x \, dx \Rightarrow v = e^x \end{matrix} \right) = x^2 e^x – 2 \left[ xe^x – \int e^x \, dx \right] = x^2 e^x – 2xe^x + 2 \int e^x \, dx = x^2 e^x – 2xe^x + 2e^x + C$$
$$\int x e^{3x} \, dx = \left( \begin{matrix} u = x \Rightarrow du = dx \\ dv = e^{3x} \, dx \Rightarrow v = \frac{1}{3}e^{3x} \end{matrix} \right) = \frac{x}{3}e^{3x} – \int \frac{1}{3}e^{3x} \, dx = \frac{x}{3}e^{3x} – \frac{1}{9}e^{3x} + C$$
$$\int x^2 \ln x \, dx = \left( \begin{matrix} u = \ln x \Rightarrow du = \frac{dx}{x} \\ dv = x^2 \, dx \Rightarrow v = \frac{x^3}{3} \end{matrix} \right) = \frac{x^3}{3} \ln x – \int \frac{x^3}{3} \frac{dx}{x} = \frac{x^3}{3} \ln x – \frac{1}{3} \int x^2 \, dx = \frac{x^3}{3} \ln x – \frac{x^3}{9} + C$$
$$\int \sqrt{x} \ln x \, dx = \left( \begin{matrix} u = \ln x \Rightarrow du = \frac{dx}{x} \\ dv = x^{1/2} \, dx \Rightarrow v = \frac{2}{3}x^{3/2} \end{matrix} \right) = \frac{2}{3}x^{3/2} \ln x – \int \frac{2}{3}x^{3/2} \frac{dx}{x} = \frac{2}{3}x^{3/2} \ln x – \frac{2}{3} \int x^{1/2} \, dx = \frac{2}{3}x^{3/2} \ln x – \frac{4}{9}x^{3/2} + C$$
$$\int \frac{\ln x}{x^2} \, dx = \left( \begin{matrix} u = \ln x \Rightarrow du = \frac{1}{x} dx \\ dv = x^{-2} \, dx \Rightarrow v = -x^{-1} \end{matrix} \right) = -\frac{\ln x}{x} – \int \left(-\frac{1}{x} \cdot \frac{1}{x}\right) \, dx = -\frac{\ln x}{x} + \int x^{-2} \, dx = -\frac{\ln x}{x} – \frac{1}{x} + C$$
$$\int x e^{-x} \, dx = \left( \begin{matrix} u = x \Rightarrow du = dx \\ dv = e^{-x} \, dx \Rightarrow v = -e^{-x} \end{matrix} \right) = -x e^{-x} – \int -e^{-x} \, dx = -x e^{-x} – e^{-x} + C$$
$$\int \frac{\ln x}{\sqrt{x}} \, dx = \left( \begin{matrix} u = \ln x \Rightarrow du = \frac{1}{x} dx \\ dv = x^{-1/2} \, dx \Rightarrow v = 2x^{1/2} \end{matrix} \right) = 2\sqrt{x} \ln x – \int \frac{2\sqrt{x}}{x} \, dx = 2\sqrt{x} \ln x – 4\sqrt{x} + C$$
$$\int (\ln x)^2 \, dx = \left( \begin{matrix} u = (\ln x)^2 \Rightarrow du = \frac{2 \ln x}{x} dx \\ dv = dx \Rightarrow v = x \end{matrix} \right) = x(\ln x)^2 – 2 \int \ln x \, dx = x(\ln x)^2 – 2(x \ln x – x) + C$$
$$\int (x^2 + 1) e^x \, dx = \left( \begin{matrix} u = x^2 + 1 \Rightarrow du = 2x \, dx \\ dv = e^x \, dx \Rightarrow v = e^x \end{matrix} \right) = (x^2 + 1)e^x – 2 \int x e^x \, dx = (x^2 + 1)e^x – 2(xe^x – e^x) + C$$
$$\int \frac{\ln x}{x^3} \, dx = \left( \begin{matrix} u = \ln x \Rightarrow du = \frac{1}{x} dx \\ dv = x^{-3} \, dx \Rightarrow v = -\frac{1}{2x^2} \end{matrix} \right) = -\frac{\ln x}{2x^2} + \int \frac{1}{2x^3} \, dx = -\frac{\ln x}{2x^2} – \frac{1}{4x^2} + C$$
$$\int (2x – 3) e^{2x} \, dx = \left( \begin{matrix} u = 2x – 3 \Rightarrow du = 2 \, dx \\ dv = e^{2x} \, dx \Rightarrow v = \frac{1}{2}e^{2x} \end{matrix} \right) = \frac{2x – 3}{2}e^{2x} – \int e^{2x} \, dx = \frac{2x – 3}{2}e^{2x} – \frac{1}{2}e^{2x} + C$$
$$\int \ln(\sqrt{x}) \, dx = \left( \begin{matrix} u = \ln(\sqrt{x}) \Rightarrow du = \frac{1}{2x} dx \\ dv = dx \Rightarrow v = x \end{matrix} \right) = x \ln(\sqrt{x}) – \int \frac{x}{2x} \, dx = x \ln(\sqrt{x}) – \frac{1}{2}x + C$$
$$\int (x+5) e^x \, dx = \left( \begin{matrix} u = x+5 \Rightarrow du = dx \\ dv = e^x \, dx \Rightarrow v = e^x \end{matrix} \right) = (x+5)e^x – \int e^x \, dx = (x+5)e^x – e^x + C = (x+4)e^x + C$$
$$\int x^{1/3} \ln x \, dx = \left( \begin{matrix} u = \ln x \Rightarrow du = \frac{1}{x} dx \\ dv = x^{1/3} \, dx \Rightarrow v = \frac{3}{4}x^{4/3} \end{matrix} \right) = \frac{3}{4}x^{4/3} \ln x – \int \frac{3}{4}x^{1/3} \, dx = \frac{3}{4}x^{4/3} \ln x – \frac{9}{16}x^{4/3} + C$$
Inmediatas Trigonométricas:
$$\int \sin(3x) \, dx = -\frac{1}{3}\cos(3x) + C$$
$$\int x \cos(x) \, dx = x \sin(x) + \cos(x) + C$$
$$\int \frac{1}{\sqrt{1-x^2}} \, dx = \sin^{-1}(x) + C$$
$$\int \frac{\tan(x)}{\cos^2(x)} \, dx = \frac{1}{2}\tan^2(x) + C \text{ o } \frac{1}{2}\sec^2(x) + C$$
$$\int (3e^x – \sin x \, dx = 3 \int e^x \, dx – \int \sin x \, dx = 3e^x + \cos x + C$$
$$\int \frac{3}{5x^2 + 5} \, dx = \frac{3}{5} \int \frac{1}{x^2 + 1} \, dx = \frac{3}{5} \arctan x + C$$
$$\int \sqrt{\frac{4}{9 – 9x^2}} \, dx = \int \sqrt{\frac{4}{9}} \sqrt{\frac{1}{1 – x^2}} \, dx = \sqrt{\frac{4}{9}} \int \frac{1}{\sqrt{1 – x^2}} \, dx = \frac{2}{3} \arcsin x + C$$
$$\int \frac{\sin x}{\sqrt{\cos x}} dx = \begin{pmatrix} t = \cos x \\ dt = -\sin x dx \end{pmatrix} = \int \frac{-dt}{\sqrt{t}} = -\int \frac{1}{t^{1/2}} dt = -\int t^{-1/2} dt = -\frac{t^{1/2}}{1/2} + C = -2t^{1/2} + C = -2\sqrt{t} + C = -2\sqrt{\cos x} + C$$
$$\int x \sin x \, dx = \left( \begin{matrix} u = x \Rightarrow du = dx \\ dv = \sin x \, dx \Rightarrow v = -\cos x \end{matrix} \right) = -x \cos x – \int (-\cos x) \, dx = -x \cos x + \int \cos x \, dx = -x \cos x + \sin x + C$$
$$\int \arctan x \, dx = \left( \begin{matrix} u = \arctan x \Rightarrow du = \frac{1}{1+x^2} dx \\ dv = dx \Rightarrow v = x \end{matrix} \right) = x \arctan x – \int \frac{x}{1+x^2} \, dx = x \arctan x – \frac{1}{2} \ln(1+x^2) + C$$
$$\int x \cos(2x) \, dx = \left( \begin{matrix} u = x \Rightarrow du = dx \\ dv = \cos(2x) \, dx \Rightarrow v = \frac{1}{2}\sin(2x) \end{matrix} \right) = \frac{x}{2}\sin(2x) – \int \frac{1}{2}\sin(2x) \, dx = \frac{x}{2}\sin(2x) + \frac{1}{4}\cos(2x) + C$$
Integrales definidas
$$\int_{0}^{2} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{2} = \frac{8}{3}$$
$$\int_{1}^{e} \frac{1}{x} \, dx = \left[ \ln|x| \right]_{1}^{e} = 1 – 0 = 1$$
$$\int_{0}^{4} \sqrt{x} \, dx = \left[ \frac{2}{3}x^{3/2} \right]_{0}^{4} = \frac{2}{3}(8) = \frac{16}{3}$$
$$\int_{1}^{3} (2x + 1) \, dx = [x^2 + x]_{1}^{3} = (9+3) – (1+1) = 10$$
$$\int_{1}^{2} \frac{1}{x^2} \, dx = [-\frac{1}{x}]_{1}^{2} = -\frac{1}{2} – (-1) = \frac{1}{2}$$
$$\int_{0}^{1} 2^x \, dx = [\frac{2^x}{\ln(2)}]_{0}^{1} = \frac{2}{\ln(2)} – \frac{1}{\ln(2)} = \frac{1}{\ln(2)}$$
$$\int_{0}^{1} (x+1)^3 \, dx = [\frac{(x+1)^4}{4}]_{0}^{1} = \frac{16}{4} – \frac{1}{4} = \frac{15}{4}$$
$$\int_{-1}^{1} (3x^2 – 1) \, dx = [x^3 – x]_{-1}^{1} = (1-1) – (-1 – (-1)) = 0$$
$$\int_{1}^{9} \frac{1}{\sqrt{x}} \, dx = [2\sqrt{x}]_{1}^{9} = 2(3) – 2(1) = 4$$
$$\int_{0}^{2} (x^3 + 3x^2 – 5x + 2) \, dx = [\frac{x^4}{4} + x^3 – \frac{5x^2}{2} + 2x]_{0}^{2} = 4 + 8 – 10 + 4 = 6$$
$$\int_{-1}^{1} (4x^3 – 2x + 6) \, dx = [x^4 – x^2 + 6x]_{-1}^{1} = (1-1+6) – (1-1-6) = 12$$
$$\int_{0}^{3} (x-2)^2 \, dx = \int_{0}^{3} (x^2 – 4x + 4) \, dx = [\frac{x^3}{3} – 2x^2 + 4x]_{0}^{3} = 9 – 18 + 12 = 3$$