I Resolver las siguientes integrales

 

1 \displaystyle \int \cfrac{1}{x^{2}\sqrt[5]{x^{2}}}\, dx

 

2 \displaystyle \int (x+2)^{3}dx

 

3 \displaystyle \int (2x+1)(x^{2}+x+1)\, dx

 

4 \displaystyle \int \cfrac{x+1}{\sqrt[3]{x^{2}+2x+7}}\, dx

 

5 \displaystyle \int \sin 2x\cos 2x\, dx

 

6 \displaystyle \int \sin^{4}x\cos x\, dx

 

7 \displaystyle \int \tan^{2}x\sec^{2}x\, dx

 

8 \displaystyle \int \cfrac{\arctan x}{1+x^{2}}\, dx

 

1 \displaystyle \int \cfrac{1}{x^{2}\sqrt[5]{x^{2}}}\, dx

\displaystyle \int \cfrac{1}{x^{2}\sqrt[5]{x^{2}}}\, dx=\int x^{-2}x^{-\frac{2}{5}}dx=\int x^{-\frac{12}{5}}dx=\cfrac{x^{-\frac{12}{5}+1}}{-\cfrac{12}{5}+1}

 

=\cfrac{x^{-\frac{7}{5}}}{-\cfrac{7}{5}}=-\cfrac{5}{7\sqrt[5]{x^{7}}}+\textup{C}

 

2 \displaystyle \int (x+2)^{3}dx

 

\displaystyle \int (x+2)^{3}dx=\cfrac{1}{4}\, (x+2)^{4}+\textup{C}

 

3 \displaystyle \int (2x+1)(x^{2}+x+1)\, dx

 

\displaystyle \int (2x+1)(x^{2}+x+1)\, dx=\cfrac{1}{2}\, (x^{2}+x+1)^{2}+\textup{C}

 

4 \displaystyle \int \cfrac{x+1}{\sqrt[3]{x^{2}+2x+7}}\, dx

 

\displaystyle \int \cfrac{x+1}{\sqrt[3]{x^{2}+2x+7}}\, dx=\cfrac{1}{2}\int (2x+2)(x^{2}+2x+7)^{-\frac{1}{3}}dx

 

=\cfrac{1}{2}\cdot \cfrac{(x^{2}+2x+7)^{\frac{2}{3}}}{\cfrac{2}{3}}=\cfrac{3}{4}\sqrt[3]{(x^{2}+2x+7)^{2}}+\textup{C}

 

5 \displaystyle \int \sin 2x\cos 2x\, dx

 

\displaystyle \int \sin 2x\cos 2x\, dx=\cfrac{1}{2}\int \sin2x\cos2x\cdot 2dx=\cfrac{1}{4}\sin^{2}2x+\textup{C}

 

6 \displaystyle \int \sin^{4}x\cos x\, dx

 

\displaystyle \int \sin^{4}x\cos x\, dx=\cfrac{1}{5}\sin^{5}x+\textup{C}

 

7 \displaystyle \int \tan^{2}x\sec^{2}x\, dx

 

\displaystyle \int \tan^{2}x\sec^{2}x\, dx=\cfrac{1}{3}\tan^{3}x+\textup{C}

 

8 \displaystyle \int \cfrac{\arctan x}{1+x^{2}}\, dx

 

\displaystyle \int \cfrac{\arctan x}{1+x^{2}}\, dx=\cfrac{1}{2}(\arctan x)^{2}+\textup{C}

 

 

Superprof

II Calcular las integrales

 

1 \displaystyle \int \cfrac{2x}{1+x^{2}}\, dx

 

2 \displaystyle \int \tan x\, dx

 

3 \displaystyle \int \cfrac{5^{3x}}{5^{3x}+7}\, \, dx

 

4 \displaystyle \int \cfrac{1}{x\ln x}\, dx

 

5 \displaystyle \int \cfrac{1}{\cos^{2}x \tan x}\, dx

 

6 \displaystyle \int \cfrac{x+1}{x}\, dx

 

7 \displaystyle \int \cfrac{x+1}{x-5}\, dx

 

8 \displaystyle \int \cfrac{3x^{3}+5x}{x^{2}+1}\, dx

 

1 \displaystyle \int \cfrac{2x}{1+x^{2}}\, dx

\displaystyle \int \cfrac{2x}{1+x^{2}}\, dx=\ln (1+x^{2})+\textup{C}

 

2 \displaystyle \int \tan x\, dx

 

\displaystyle \int \tan x\, dx=\int \cfrac{\sin x}{\cos x}\, dx=-\int \frac{-\sin x}{\cos x}=-\ln \cos x+\textup{C}

 

3 \displaystyle \int \cfrac{5^{3x}}{5^{3x}+7}\, \, dx

 

\displaystyle \int \cfrac{5^{3x}}{5^{3x}+7}\, \, dx=\cfrac{1}{3}\cdot \cfrac{1}{\ln 5}\int \cfrac{3\cdot 5^{3x}\ln 5}{5^{3x}+7}\, dx=\cfrac{1}{3\ln 5}\, \ln (5^{3x}+7)+\textup{C}

 

4 \displaystyle \int \cfrac{1}{x\ln x}\, dx

 

\displaystyle \int \cfrac{1}{x\ln x}\, dx=\int \cfrac{\cfrac{1}{x}}{\ln x}\, dx=\ln (\ln x)+\textup{C}

 

5 \displaystyle \int \cfrac{1}{\cos^{2}x \tan x}\, dx

 

\displaystyle \int \cfrac{1}{\cos^{2}x \tan x}\, dx=\int \cfrac{\sec^{2}x }{\tan x}\, dx=\ln (\tan x)+\textup{C}

 

6 \displaystyle \int \cfrac{x+1}{x}\, dx

 

\displaystyle \int \cfrac{x+1}{x}\, dx=\int \left ( 1+\cfrac{1}{x} \right )dx=x+\ln x+\textup{C}

 

7 \displaystyle \int \cfrac{x+1}{x-5}\, dx

 

\displaystyle \int \cfrac{x+1}{x-5}\, dx=\int \cfrac{x+1-5+5}{x-5}\, dx=\int \cfrac{x-5}{x-5}\, dx+\int \cfrac{6}{x-6}\, dx

 

=x+6 \ln (x-5)+\textup{C}

 

8 \displaystyle \int \cfrac{3x^{3}+5x}{x^{2}+1}\, dx

 

\begin{matrix} 3x^{3}+5x\\ \; \; \; \; \; \; \; \; \; 2x \end{matrix}\; \; \; \begin{matrix} \underline{\mid x^{2}+1}\\ 3x \end{matrix}

 

\displaystyle \int \cfrac{3x^{3}+5x}{x^{2}+1}\, dx=\int \left ( 3x+\cfrac{2x}{x^{2}+1} \right )dx=\cfrac{3}{2}\, x^{2}+\ln (x^{2}+1)+\textup{C}

 

III Resolver las siguientes integrales exponenciales

 

1 \displaystyle \int e^{2x+2}dx

 

2 \displaystyle \int 5^{x}dx

 

3 \displaystyle \int 2^{x}\cdot 5^{x}dx

 

4 \displaystyle \int 8^{3x+1}dx

 

5 \displaystyle \int \cfrac{e^{\ln x}}{x}\, dx

 

6 \displaystyle \int e^{\sin x}\cos x\, dx

 

7 \displaystyle \int \cfrac{e^{\arcsin x}}{\sqrt{1-x^{2}}}\, dx

 

1 \displaystyle \int e^{2x+2}dx

\displaystyle \int e^{2x+2}dx=\frac{1}{2}\, e^{2x+2}+\textup{C}

 

2 \displaystyle \int 5^{x}dx

 

\displaystyle \int \cfrac{5^{x}}{\ln 5}+\textup{C}

 

3 \displaystyle \int 2^{x}\cdot 5^{x}dx

 

\displaystyle \int 2^{x}\cdot 5^{x}dx=\int 10^{x}dx=\frac{10^{x}}{\ln 10}+\textup{C}

 

4 \displaystyle \int 8^{3x+1}dx

 

\displaystyle \int 8^{3x+1}dx=\cfrac{1}{3}\int 8^{3x+1}\cdot 3\, dx=\cfrac{1}{3\ln 8}\cdot 8^{3x+1}+\textup{C}

 

5 \displaystyle \int \cfrac{e^{\ln x}}{x}\, dx

 

\displaystyle \int \cfrac{e^{\ln x}}{x}\, dx=\int \cfrac{1}{x}\, e^{\ln x}dx=e^{\ln x}+\textup{C}

 

6 \displaystyle \int e^{\sin x}\cos x\, dx

 

\displaystyle \int e^{\sin x}\cos x\, dx=e^{\sin x}+\textup{C}

 

7 \displaystyle \int \cfrac{e^{\arcsin x}}{\sqrt{1-x^{2}}}\, dx

 

\displaystyle \int \cfrac{e^{\arcsin x}}{\sqrt{1-x^{2}}}\, dx=\int \cfrac{1}{\sqrt{1-x^{2}}}\, e^{\arcsin x}dx=e^{\arcsin x}+\textup{C}

 

IV  Calcular las integrales

 

1 \displaystyle \int (3-\sin x)dx

 

2 \displaystyle \int \sin (3x+5)\, dx

 

3 \displaystyle \int (x+1)\sin (x^{2}+2x+3)\, dx

 

4 \displaystyle \int e^{x}\sin e^{x}\, dx

 

5 \displaystyle \int \sin 2x\, dx

 

6 \displaystyle \int \sin^{2}2x\, dx

 

7 \displaystyle \int \sin^{3}x\, dx

 

1 \displaystyle \int (3-\sin x)dx

\displaystyle \int (3-\sin x)dx=3x+\cos x+\textup{C}

 

2 \displaystyle \int \sin (3x+5)\, dx

 

\displaystyle \int \sin (3x+5)\, dx=\cfrac{1}{3}\int \sin (3x+5)\cdot 3\, dx=-\cfrac{1}{3}\cos (3x+5)+\textup{C}

 

3 \displaystyle \int (x+1)\sin (x^{2}+2x+3)\, dx

 

\displaystyle \int (x+1)\sin (x^{2}+2x+3)\, dx=\cfrac{1}{2}\int (2x+2)\sin (x^{2}+2x+3)\, dx

 

=-\frac{1}{2}\cos (x^{2}+2x+3)+\textup{C}

 

4 \displaystyle \int e^{x}\sin e^{x}\, dx

 

\displaystyle \int e^{x}\sin e^{x}\, dx=-\cos e^{x}+\textup{C}

 

5\displaystyle \int \sin 2x\, dx

 

\displaystyle \int \sin 2x\, dx=\cfrac{1}{2}\int \sin 2x\cdot 2\, dx=-\cfrac{1}{2}\cos 2x +\textup{C}

 

6 \displaystyle \int \sin^{2}2x\, dx

 

\displaystyle \int \sin^{2}2x\, dx=\cfrac{1-4\cos 4x}{2}\, dx=\cfrac{1}{2}\, x-\cfrac{1}{8}\sin 4x +\textup{C}

 

7 \displaystyle \int \sin^{3}x\, dx

 

\displaystyle \int \sin^{3}x\, dx=\int \sin^{2}x\cdot \sin x\, dx=\int (1-\cos^{2}x)\sin x\, dx

 

\displaystyle =\int (\sin x-\cos^{2}x\sin x)\, dx=-\cos x +\cfrac{1}{3}\cos^{3}x+\textup{C}

 

 

V Resolver las integrales

 

1 \displaystyle =\int (2x+\cos x)dx

 

2 \displaystyle =\int \cos (2x+5)dx

 

3 \displaystyle =\int (x+1)\cos (x^{2}+2x+1)dx

 

4 \displaystyle \int \cfrac{\cos (\ln x)}{x}\, dx

 

5 \displaystyle \int \cos^{2}x\, dx

 

6 \displaystyle \int \cos^{3}3x\, dx

 

1 \displaystyle =\int (2x+\cos x)dx

\displaystyle =\int (2x+\cos x)dx=x^{2}+\sin x +\textup{C}

 

2 \displaystyle =\int \cos (2x+5)dx

 

\displaystyle =\int \cos (2x+5)dx=\cfrac{1}{2}\sin (2x+5)+\textup{C}

 

3 \displaystyle =\int (x+1)\cos (x^{2}+2x+1)dx

 

\displaystyle =\int (x+1)\cos (x^{2}+2x+1)dx=\cfrac{1}{2}\int (2x+2)\cos (x^{2}+2x+1)dx

 

=\cfrac{1}{2}\sin (x^{2}+2x+1)+\textup{C}

 

4 \displaystyle \int \cfrac{\cos (\ln x)}{x}\, dx

 

\displaystyle \int \cfrac{\cos (\ln x)}{x}\, dx=\int \cos (\ln x)\cdot \cfrac{1}{x}\, dx=\sin (\ln x)+\textup{C}

 

5 \displaystyle \int \cos^{2}x\, dx

 

\displaystyle \int \cos^{2}x\, dx=\int \left (\sqrt{\cfrac{1+\cos 2x}{2} \right )^{2}}\, dx=\int \cfrac{1+\cos 2x}{2}\, dx

 

\displaystyle \cfrac{1}{2}\int (1+\cos 2x)dx=\cfrac{1}{2}\left ( x+\cfrac{1}{2}\sin 2x \right )=\cfrac{1}{2}\, x+\cfrac{1}{4}\sin 2x+\textup{C}

 

6 \displaystyle \int \cos^{3}3x\, dx

 

\displaystyle \int \cos^{3}3x\, dx =\int \cos^{2}3x\cos 3x\, dx=\int (1-\sin^{2}3x)\cos 3x\, dx

 

\displaystyle \int \cos 3x\, dx-\int \sin^{2}3x\cos 3x\, dx=\cfrac{1}{3}\sin 3x-\frac{1}{9}\sin^{3}3x+\textup{C}

 

VI Calcular las integrales

 

1 \displaystyle \int \cfrac{5}{\cos^{2}x}\, dx

 

2 \displaystyle \int (3+3\tan^{2}x)dx

 

3 \displaystyle \int \sec^{2}(5x+3)dx

 

4 \displaystyle \int \sec^{4}x\, dx

 

5 \displaystyle \int (3+3\cot^{2}x)dx

 

6 \displaystyle \int \tan^{2}x\, dx

 

1 \displaystyle \int \cfrac{5}{\cos^{2}x}\, dx

\displaystyle \int \cfrac{5}{\cos^{2}x}\, dx=5\tan x+\textup{C}

 

2 \displaystyle \int (3+3\tan^{2}x)dx

 

\displaystyle \int (3+3\tan^{2}x)dx=3\int (1+\tan^{2}x)dx=3\tan x+\textup{C}

 

3 \displaystyle \int \sec^{2}(5x+3)dx

 

\displaystyle \int \sec^{2}(5x+3)dx=\cfrac{1}{5}\int \sec^{2}(5x+3)\cdot 5\, dx=\cfrac{1}{5}\tan (5x+3)+\textup{C}

 

4 \displaystyle \int \sec^{4}x\, dx

 

\displaystyle \int \sec^{4}x\, dx=\int \sec^{2}x \sec^{2}x\, dx=\int (1+\tan^{2}x)\sec^{2}x\, dx

 

\displaystyle \int (\sec^{2}x+\sec^{2}x\tan^{2}x)dx=\tan x +\cfrac{1}{3}\tan^{3}x+\textup{C}

 

5 \displaystyle \int (3+3\cot^{2}x)dx

 

\displaystyle \int (3+3\cot^{2}x)dx=3\int (1+ \cot^{2}x)dx=-3\cot x+\textup{C}

 

6 \displaystyle \int \tan^{2}x\, dx

 

\displaystyle \int \tan^{2}x\, dx=\int (1+\tan^{2}x-1)dx=\int (1+\tan^{2}x)dx-\int dx=\tan x -x+\textup{C}

 

VII Resolver las integrales

 

1 \displaystyle \int \sin 3x\cos 2x\, dx

 

2 \displaystyle \int \cfrac{dx}{\sin^{2}x\cos^{2}x}

 

3 \displaystyle \int \sqrt{\cfrac{1+x}{1-x}}\, dx

 

4 \displaystyle \int \cfrac{1-\cos x}{1+\cos x}\, dx

 

5 \displaystyle \int \cfrac{1+\sin x}{1-\sin x}\, dx

 

1 \displaystyle \int \sin 3x\cos 2x\, dx

\sin A+\sin B=2\sin \cfrac{A+B}{2}\cdot \cos \cfrac{A-B}{2}

 

\left\{\begin{matrix} \cfrac{A+B}{2}=3x\\ \\ \cfrac{A-B}{2}=2x \end{matrix}\right. \; \; \; \; \; \; \; \; \; \; A=5x\; \; \; \; \; B=x

 

\displaystyle \int \sin 3x\cdot \cos 2x\, dx=\cfrac{1}{2}\int 2\sin 3x\cdot \cos 2x\, dx

 

\displaystyle=\cfrac{1}{2}\int (\sin 5x+\sin x)dx=\cfrac{1}{2}\left ( -\cfrac{\cos 5x}{5}-\cos x \right )+\textup{C}

 

2 \displaystyle \int \cfrac{dx}{\sin^{2}x\cos^{2}x}

 

\displaystyle \int \cfrac{dx}{\sin^{2}x\cos^{2}x}=\int \cfrac{\sin^{2}x+\cos^{2}x}{\sin^{2}x\cos^{2}x}\, dx

 

\displaystyle \int \cfrac{\sin^{2}x}{\sin^{2}x\cos^{2}x}\, dx+\int \cfrac{\cos^{2}x}{\sin^{2}x\cos^{2}x}\, dx

 

\displaystyle \int \cfrac{dx}{\cos^{2}x}+\int \cfrac{dx}{\sin^{2}x}=\tan x-\cot x+\textup{C}

 

3 \displaystyle \int \sqrt{\cfrac{1+x}{1-x}}\, dx

 

\displaystyle \int \sqrt{\cfrac{1+x}{1-x}}\, dx=\int \cfrac{1-x}{\sqrt{1-x^{2}}}\, dx=\int \cfrac{1}{\sqrt{1-x^{2}}}\, dx+\int \cfrac{x}{\sqrt{1-x^{2}}}\, dx

 

\displaystyle \int \cfrac{1}{\sqrt{1-x^{2}}}\, dx -\cfrac{1}{2}\int (-2x)(1-x^{2})^{-\frac{1}{2}}dx

 

\arcsin x-\sqrt{1-x^{2}}+\textup{C}

 

4 \displaystyle \int \cfrac{1-\cos x}{1+\cos x}\, dx

 

\displaystyle \int \cfrac{1-\cos x}{1+\cos x}\, dx=\int \cfrac{(1-\cos x)^{2}}{1-\cos^{2}x}\, dx=\int \cfrac{1-2\cos x+\cos^{2}x}{\sin^{2}x}\, dx

 

\displaystyle \int \cfrac{1}{\sin^{2}x}\, dx-2\int \cos x\sin^{-2}x\, dx+\int \cot^{2}x\, dx

 

\displaystyle \int \cfrac{1}{\sin^{2}x}\, dx-2\int \cos x\sin^{-2}x\, dx+\int \left [\left ( 1+\cot ^{2}x \right )-1 \right ] dx

 

=-\cot x +\cfrac{2}{\sin x}-\cot x-x

 

=-2\cot x +\cfrac{2}{\sin x}-x+\textup{C}

 

5 \displaystyle \int \cfrac{1+\sin x}{1-\sin x}\, dx

 

\displaystyle \int \cfrac{1+\sin x}{1-\sin x}\, dx=\int \cfrac{1+2\sin x +\sin^{2}x}{\cos^{2}x}\, dx

 

\displaystyle =\int \cfrac{1}{\cos^{2}x}\, dx+2\int \cfrac{\sin x}{\cos^{2}x}\, dx+\int \cfrac{\sin^{2}x}{\cos^{2}x}\, dx

 

\displaystyle =\int \cfrac{1}{\cos^{2}x}\, dx+2\int (-\sin x)\cos^{-2}x\, dx+\int \tan^{2}x\,dx

 

\displaystyle =\int \cfrac{1}{\cos^{2}x}\, dx+2\int (-\sin x)\cos^{-2}x\, dx+\int \left [\left (1+\tan^{2}x \right )-1 \right ]dx

 

=\tan x+\cfrac{2}{\cos x}+\tan x-x=2\tan x+\cfrac{2}{\cos x}-x+\textup{C}

 

VIII Calcular las integrales

 

1 \displaystyle \int \cfrac{5}{x^{2}-4x+8}\, dx

2 \displaystyle \int \cfrac{2x+5}{\sqrt{9-x^{2}}}\, dx

3 \displaystyle \int \cfrac{2^{x}}{\sqrt{1-4^{x}}}\, dx

4 \displaystyle \int \cfrac{x}{\sqrt{9-2x^{4}}}\, dx

 

1 \displaystyle \int \cfrac{5}{x^{2}-4x+8}\, dx

\displaystyle \int \cfrac{5}{x^{2}-4x+8}\, dx=\int \cfrac{5}{x^{2}-4x+4+4}\, dx=\int \cfrac{5}{4+(x-2)^{2}}\, dx

 

\displaystyle = \frac{5}{4} \int \cfrac{dx}{1+\left ( \cfrac{x-2}{2} \right )^{2}}=\cfrac{5}{4}\cdot 2\int \cfrac{\cfrac{1}{2}}{1+\left ( \cfrac{x-2}{2} \right )^{2}}\, dx

 

=\cfrac{5}{2}\arctan \left ( \cfrac{x-2}{2} \right )+\textup{C}

 

2 \displaystyle \int \cfrac{2x+5}{\sqrt{9-x^{2}}}\, dx

 

\displaystyle = \int \cfrac{2x+5}{\sqrt{9-x^{2}}}\, dx=\int \cfrac{2x}{\sqrt{9-x^{2}}}\, dx+\int \cfrac{5}{\sqrt{9-x^{2}}}\, dx

 

\displaystyle =-\int (9-x^{2})^{-\frac{1}{2}}(-2x)dx+\cfrac{5}{3}\cdot 3\int \cfrac{\cfrac{1}{3}}{\sqrt{1-\left ( \cfrac{x}{3} \right )^{2}}}\, dx

 

=-\cfrac{(9-x^{2})^{\frac{1}{2}}}{\cfrac{1}{2}}+5\arcsin \left ( \cfrac{x}{3} \right )=-2\sqrt{9-x^{2}}+5\arcsin \left ( \cfrac{x}{3} \right )+\textup{C}

 

3 \displaystyle \int \cfrac{2^{x}}{\sqrt{1-4^{x}}}\, dx

 

\displaystyle \int \cfrac{2^{x}}{\sqrt{1-4^{x}}}\, dx=\int \cfrac{2^{x}}{\sqrt{1-(2^{x})^{2}}}\, dx=\cfrac{1}{\ln 2}\int \cfrac{2^{x}\ln 2}{\sqrt{1-(2^{x})^{2}}}\, dx

 

=\cfrac{1}{\ln 2}\arcsin (2^{x})+\textup{C}

 

4 \displaystyle \int \cfrac{x}{\sqrt{9-2x^{4}}}\, dx

 

\displaystyle \int \cfrac{x}{\sqrt{9-2x^{4}}}\, dx=\int \cfrac{x}{\sqrt{9\left [ 1-\left ( \cfrac{\sqrt{2}x^{2}}{3} \right )^{2} \right ]}}\, dx

 

\displaystyle =\cfrac{1}{3}\cdot \cfrac{3}{2\sqrt{2}}\int \cfrac{\cfrac{2\sqrt{2}}{3}\, x}{\sqrt{1-\left ( \cfrac{\sqrt{2}\, x^{2}}{3} \right )^{2}}}\, dx=\cfrac{1}{2\sqrt{2}}\arcsin \left ( \cfrac{\sqrt{2}}{3}\, x^{2} \right )+\textup{C}

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  S’abonner  
Notifier de
ALDO DE LEON
ALDO DE LEON
Invité
17 May.

¿cuál es la integral por partes de esta ecuación?

xquinta coseno(3x)dx

x3cubica e3/4dx

No lo entiendo, sustituyo valores… ¿en cual debo de integrar?

Zhao
Zhao
Invité
30 May.

en ejercicio 2 de VI puede que hayas solucionado mal.
int(3+tg^2x)dx

Karla Paulette Flores Silva
Karla Paulette Flores Silva
Editor
28 Jun.

Hola, muchas gracias por tu comentario. Efectivamente había un error pero ya lo hemos corregido

¡saludos!