Adintyoko 2404: Mei 2015

Berikut ini adalah rumus-rumus dasar integral taktentu:

$\int \,{\rm d}x = x + C$
$\int x^n\,{\rm d}x = \frac{x^{n+1}}{n+1} + C\qquad\mbox{ jika }n \ne -1$
$\int {dx \over x} = \ln{\left|x\right|} + C$
$\int {dx \over {a^2+x^2}} = {1 \over a}\arctan {x \over a} + C$
$\int e^x\,dx = e^x + C$
$\int a^x\,dx = \frac{a^x}{\ln{a}} + C$
$\int \sin{x}\, dx = -\cos{x} + C$
$\int \cos{x}\, dx = \sin{x} + C$
$\int \tan{x} \, dx = \ln{\left| \sec {x} \right|} + C$
$\int \cot{x} \, dx = -\ln{\left| \csc{x} \right|} + C$
$\int \sec{x} \, dx = \ln{\left| \sec{x} + \tan{x}\right|} + C$
$\int \csc{x} \, dx = -\ln{\left| \csc{x} + \cot{x}\right|} + C$
$\int \sec^2 x \, dx = \tan x + C$
$\int \csc^2 x \, dx = -\cot x + C$
$\int \sec{x} \, \tan{x} \, dx = \sec{x} + C$
$\int \csc{x} \, \cot{x} \, dx = - \csc{x} + C$
$\int \sin^2 x \, dx = \frac{1}{2}(x - \sin x \cos x) + C$
$\int \cos^2 x \, dx = \frac{1}{2}(x + \sin x \cos x) + C$
$\int \sec^3 x \, dx = \frac{1}{2}\sec x \tan x + \frac{1}{2}\ln|\sec x + \tan x| + C$
$\int \sin^n x \, dx = - \frac{\sin^{n-1} {x} \cos {x}}{n} + \frac{n-1}{n} \int \sin^{n-2}{x} \, dx$
$\int \cos^n x \, dx = \frac{\cos^{n-1} {x} \sin {x}}{n} + \frac{n-1}{n} \int \cos^{n-2}{x} \, dx$
$\int \arctan{x} \, dx = x \, \arctan{x} - \frac{1}{2} \ln{\left| 1 + x^2\right|} + C$

Berikut adalah rumus-rumus dasar turunan:

c' = 0 \,

x' = 1 \,

(cx)' = c \,

|x|' = {x \over |x|} = \sgn x, \qquad x \ne 0

(x^c)' = cx^{c-1} \qquad \mbox{baik } x^c \mbox{ maupun } cx^{c-1} \mbox { terdefinisi}

\left({1 \over x}\right)' = \left(x^{-1}\right)' = -x^{-2} = -{1 \over x^2}

\left({1 \over x^c}\right)' = \left(x^{-c}\right)' = -cx^{-(c+1)} = -{c \over x^{c+1}}

\left(\sqrt{x}\right)' = \left(x^{1\over 2}\right)' = {1 \over 2} x^{-{1\over 2}} = {1 \over 2 \sqrt{x}}, \qquad x > 0

\left(c^x\right)' = c^x \ln c ,\qquad c > 0

\left(e^x\right)' = e^x

\left(ln x\right)' = \frac{1}{x}

Adintyoko 2404