Inicio › Matemáticas 5 › Integrales indefinidas

Matemáticas 5 · Tema 14

Integrales indefinidas

Apuntes — Formulario de integración

Constante
∫k dx=kx+c\displaystyle \int k \, dx = kx + c
Potencia
∫xn dx=xn+1n+1+c\displaystyle \int x^n \, dx = \dfrac{x^{n+1}}{n+1} + c
Uno entre x
∫1x dx=ln⁡∣x∣+c\displaystyle \int \dfrac{1}{x} \, dx = \ln|x| + c
Seno
∫sen⁡x dx=−cos⁡x+c\displaystyle \int \operatorname{sen} x \, dx = -\cos x + c
Coseno
∫cos⁡x dx=sen⁡x+c\displaystyle \int \cos x \, dx = \operatorname{sen} x + c
Secante al cuadrado
∫sec⁡2x dx=tan⁡x+c\displaystyle \int \sec^2 x \, dx = \tan x + c
Exponencial exe^x
∫ex dx=ex+c\displaystyle \int e^x \, dx = e^x + c
Exponencial axa^x
∫ax(ln⁡a) dx=ax+c\displaystyle \int a^x (\ln a) \, dx = a^x + c

Ejemplo

Se integra término a término, cada uno con su fórmula del formulario:

∫(5x2+4sen⁡x−2+3sec⁡2x+5x+7ex)dx==5x33−4cos⁡x−2x+3tan⁡x+5ln⁡∣x∣+7ex+c\begin{aligned} &\int \left( 5x^2 + 4\operatorname{sen} x - 2 + 3\sec^2 x + \frac{5}{x} + 7e^x \right) dx = \\ &= \mathbf{\frac{5x^3}{3} - 4\cos x - 2x + 3\tan x + 5\ln|x| + 7e^x + c} \end{aligned}

Ejercicios

Bloque 1
∫(4sec⁡2x−8x+6−3sen⁡x+4ex−6x4)dx\displaystyle \int \left( 4\sec^2 x - 8x + 6 - 3\operatorname{sen} x + 4e^x - 6x^4 \right) dx
∫(−4cos⁡x−5+3x−2xln⁡2−8x3+4ex)dx\displaystyle \int \left( -4\cos x - 5 + \frac{3}{x} - 2^x \ln 2 - 8x^3 + 4e^x \right) dx
∫(5sen⁡x−4−7x4−5x+7sec⁡2x−5x)dx\displaystyle \int \left( 5\operatorname{sen} x - 4 - 7x^4 - \frac{5}{x} + 7\sec^2 x - 5x \right) dx
Bloque 2
∫(6−7x2−8x+4cos⁡x−7xln⁡7−ex)dx\displaystyle \int \left( 6 - 7x^2 - \frac{8}{x} + 4\cos x - 7^x \ln 7 - e^x \right) dx
∫(2sen⁡x−9x2+6x3−3x+2ex−4cos⁡x)dx\displaystyle \int \left( 2\operatorname{sen} x - 9x^2 + 6x^3 - \frac{3}{x} + 2e^x - 4\cos x \right) dx
∫(7x2−4x+3x−2+4ex−5sen⁡x−sec⁡2x)dx\displaystyle \int \left( 7x^2 - 4x + \frac{3}{x} - 2 + 4e^x - 5\operatorname{sen} x - \sec^2 x \right) dx

Respuestas

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Bloque 1
∫(4sec⁡2x−8x+6−3sen⁡x+4ex−6x4)dx==4tan⁡x−8x22+6x−3(−cos⁡x)+4ex−6x55+c==4tan⁡x−4x2+6x+3cos⁡x+4ex−6x55+c\begin{aligned} &\int \left( 4\sec^2 x - 8x + 6 - 3\operatorname{sen} x + 4e^x - 6x^4 \right) dx = \\ &= 4\tan x - \frac{8x^2}{2} + 6x - 3(-\cos x) + 4e^x - \frac{6x^5}{5} + c = \\ &= \mathbf{4\tan x - 4x^2 + 6x + 3\cos x + 4e^x - \frac{6x^5}{5} + c} \end{aligned}
∫(−4cos⁡x−5+3x−2xln⁡2−8x3+4ex)dx==−4sen⁡x−5x+3ln⁡∣x∣−2x−8x44+4ex+c==−4sen⁡x−5x+3ln⁡∣x∣−2x−2x4+4ex+c\begin{aligned} &\int \left( -4\cos x - 5 + \frac{3}{x} - 2^x \ln 2 - 8x^3 + 4e^x \right) dx = \\ &= -4\operatorname{sen} x - 5x + 3\ln|x| - 2^x - \frac{8x^4}{4} + 4e^x + c = \\ &= \mathbf{-4\operatorname{sen} x - 5x + 3\ln|x| - 2^x - 2x^4 + 4e^x + c} \end{aligned}
∫(5sen⁡x−4−7x4−5x+7sec⁡2x−5x)dx==5(−cos⁡x)−4x−7x55−5ln⁡∣x∣+7tan⁡x−5x22+c==−5cos⁡x−4x−7x55−5ln⁡∣x∣+7tan⁡x−5x22+c\begin{aligned} &\int \left( 5\operatorname{sen} x - 4 - 7x^4 - \frac{5}{x} + 7\sec^2 x - 5x \right) dx = \\ &= 5(-\cos x) - 4x - \frac{7x^5}{5} - 5\ln|x| + 7\tan x - \frac{5x^2}{2} + c = \\ &= \mathbf{-5\cos x - 4x - \frac{7x^5}{5} - 5\ln|x| + 7\tan x - \frac{5x^2}{2} + c} \end{aligned}
Bloque 2
∫(6−7x2−8x+4cos⁡x−7xln⁡7−ex)dx==6x−7x33−8ln⁡∣x∣+4sen⁡x−7x−ex+c\begin{aligned} &\int \left( 6 - 7x^2 - \frac{8}{x} + 4\cos x - 7^x \ln 7 - e^x \right) dx = \\ &= \mathbf{6x - \frac{7x^3}{3} - 8\ln|x| + 4\operatorname{sen} x - 7^x - e^x + c} \end{aligned}
∫(2sen⁡x−9x2+6x3−3x+2ex−4cos⁡x)dx==2(−cos⁡x)−9x33+6x44−3ln⁡∣x∣+2ex−4sen⁡x+c==−2cos⁡x−3x3+3x42−3ln⁡∣x∣+2ex−4sen⁡x+c\begin{aligned} &\int \left( 2\operatorname{sen} x - 9x^2 + 6x^3 - \frac{3}{x} + 2e^x - 4\cos x \right) dx = \\ &= 2(-\cos x) - \frac{9x^3}{3} + \frac{6x^4}{4} - 3\ln|x| + 2e^x - 4\operatorname{sen} x + c = \\ &= \mathbf{-2\cos x - 3x^3 + \frac{3x^4}{2} - 3\ln|x| + 2e^x - 4\operatorname{sen} x + c} \end{aligned}
∫(7x2−4x+3x−2+4ex−5sen⁡x−sec⁡2x)dx==7x33−4x22+3ln⁡∣x∣−2x+4ex−5(−cos⁡x)−tan⁡x+c==7x33−2x2+3ln⁡∣x∣−2x+4ex+5cos⁡x−tan⁡x+c\begin{aligned} &\int \left( 7x^2 - 4x + \frac{3}{x} - 2 + 4e^x - 5\operatorname{sen} x - \sec^2 x \right) dx = \\ &= \frac{7x^3}{3} - \frac{4x^2}{2} + 3\ln|x| - 2x + 4e^x - 5(-\cos x) - \tan x + c = \\ &= \mathbf{\frac{7x^3}{3} - 2x^2 + 3\ln|x| - 2x + 4e^x + 5\cos x - \tan x + c} \end{aligned}

Ejercicios extra 1

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Bloque A
∫(3sec⁡2x−5x+2−4sen⁡x+6ex−10x3)dx\displaystyle \int \left( 3\sec^2 x - 5x + 2 - 4\operatorname{sen} x + 6e^x - 10x^3 \right) dx
∫(−6cos⁡x−2+4x−3xln⁡3−5x4+2ex)dx\displaystyle \int \left( -6\cos x - 2 + \frac{4}{x} - 3^x \ln 3 - 5x^4 + 2e^x \right) dx
∫(8sen⁡x−7−3x2−2x+5sec⁡2x−4x)dx\displaystyle \int \left( 8\operatorname{sen} x - 7 - 3x^2 - \frac{2}{x} + 5\sec^2 x - 4x \right) dx
∫(5−4x3−7x+2cos⁡x−5xln⁡5−ex)dx\displaystyle \int \left( 5 - 4x^3 - \frac{7}{x} + 2\cos x - 5^x \ln 5 - e^x \right) dx
∫(3sen⁡x−6x2+8x3−2x+5ex−2cos⁡x)dx\displaystyle \int \left( 3\operatorname{sen} x - 6x^2 + 8x^3 - \frac{2}{x} + 5e^x - 2\cos x \right) dx
∫(5x2−6x+2x−1+3ex−4sen⁡x−sec⁡2x)dx\displaystyle \int \left( 5x^2 - 6x + \frac{2}{x} - 1 + 3e^x - 4\operatorname{sen} x - \sec^2 x \right) dx
Bloque B
∫(2sec⁡2x−9x2+5−2sen⁡x+3ex−7x4)dx\displaystyle \int \left( 2\sec^2 x - 9x^2 + 5 - 2\operatorname{sen} x + 3e^x - 7x^4 \right) dx
∫(−3cos⁡x−4+6x−4xln⁡4−9x2+5ex)dx\displaystyle \int \left( -3\cos x - 4 + \frac{6}{x} - 4^x \ln 4 - 9x^2 + 5e^x \right) dx
∫(6sen⁡x−1−2x4−3x+4sec⁡2x−2x)dx\displaystyle \int \left( 6\operatorname{sen} x - 1 - 2x^4 - \frac{3}{x} + 4\sec^2 x - 2x \right) dx
∫(9−3x2−4x+5cos⁡x−6xln⁡6−2ex)dx\displaystyle \int \left( 9 - 3x^2 - \frac{4}{x} + 5\cos x - 6^x \ln 6 - 2e^x \right) dx
∫(4sen⁡x−12x2+10x3−5x+6ex−3cos⁡x)dx\displaystyle \int \left( 4\operatorname{sen} x - 12x^2 + 10x^3 - \frac{5}{x} + 6e^x - 3\cos x \right) dx
∫(8x2−2x+4x−3+2ex−3sen⁡x−2sec⁡2x)dx\displaystyle \int \left( 8x^2 - 2x + \frac{4}{x} - 3 + 2e^x - 3\operatorname{sen} x - 2\sec^2 x \right) dx
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