Common Integral Forms

I. $x^n$

We can find the integral of $x^n$ by using anti-derivatives.

$\dfrac{d}{dx} x^n = (n+1)x^n$

$\Rightarrow \displaystyle \int x^n dx = \dfrac{x^{ n+1}}{n+1}+ C$

II. $x^{-1}$

We saw that $\dfrac{1}{x}$ is the derivate of $\ln x.$

Therefore,

$\dfrac{d}{dx} \ln x = \dfrac{1}{x} $

$\Rightarrow \displaystyle \int \dfrac{1}{x} dx = \ln x + C$

III. $(ax+b)^n$

$(ax+b)^n$ is the general form of $x^n$ (where $a=1$ and $b=0$), and can be integrated the same way.

$\dfrac{d}{dx} (ax+b)^{n+1} = (n+1)(a)(ax+b)^n$

$\Rightarrow \displaystyle \int (ax+b)^n dx = \dfrac{(ax+b)^{n+1}}{(n+1)(a)} + C$

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IV. $\dfrac{1}{(ax+b)}$

$\dfrac{1}{(ax+b)}$ is also a general form of the aformentioned $\dfrac{1}{x}$ where (where $a = 1$ and $b=0$) and can be integrated likewise.

$\dfrac{d}{dx} \ln (ax+b) = \dfrac{a}{(ax+b)}$

$\Rightarrow \displaystyle \int \dfrac{1}{(ax+b)} dx = \dfrac{1}{a}\ln (ax+b) + C$

V. Other Standard Integrals Using Anti-Derivatives

i) $\displaystyle \int e^x \ dx = e^x + C$

ii) $\displaystyle \int a^x \ dx = \dfrac{a^x}{\ln a} + C$

iii) $\displaystyle \int \cos x \ dx = \sin x + C$

iv) $\displaystyle \int \sin x \ dx = - \cos x + C$

v) $\displaystyle \int \sec^2 x\ dx = \tan x + C$

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vi) $\displaystyle \int \csc x \ dx = - \cot x + C$

vii) $\displaystyle \int \dfrac{\sin x}{\cos^2 x} \ dx = \sec x + C$

viii) $\displaystyle \int \dfrac{\cos x}{\sin ^ 2 x} \ dx = \csc x + C$

VI. A Handy Trick

It is especially useful to note the following results.

$\dfrac{d}{dx} [\phi (x)]^{n+1} = (n+1)[\phi(x)]^n \phi ' (x)$

$\Rightarrow \displaystyle \int [\phi (x)]^n \phi ' (x) dx = \dfrac{[\phi (x)]^{n+1}}{n+1}$

$\dfrac{d}{dx} \ln \phi (x) = \dfrac{\phi ' (x)}{\phi x}$

$\Rightarrow \displaystyle \int \dfrac{\phi ' (x)}{\phi x} dx = \ln \phi (x)$

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VII. Common Integral Forms By Solving For Equations:

$\displaystyle \int\limits_{a}^{b} \dfrac{f(x)}{f(a + b - x) + f(x)} dx = \dfrac{b - a}{2}$

Proof:

Let $I = \displaystyle \int\limits_{a}^{b} \dfrac{f(x)}{f(a + b - x) + f(x)} dx$ $...eqn(i)$

If we substitute $u = a + b - x$, we get $du = -dx$, and for $x = a$,$u = b$ and for $x = b$, $u = a$

$\therefore I = \displaystyle \int\limits_{b}^{a} \dfrac{f(a + b - u)}{f(u) + f(a + b - u)} (-du)$

$\Rightarrow I = - \displaystyle \int\limits_{b}^{a} \dfrac{f(a + b - u)}{f(u) + f(a + b - u)} du$

$\Rightarrow I = \displaystyle \int\limits_{a}^{b} \dfrac{f(a + b - u)}{f(u) + f(a + b - u)} du$

Now, in a definite integral the variable name is insignificant, therefore, we can substitute $u = x$ in the above equation, and get:

$\Rightarrow I = \displaystyle \int\limits_{a}^{b} \dfrac{f(a + b - x)}{f(x) + f(a + b - x)} dx$ $...eqn(ii)$

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Adding equations $(i)$ and $(ii)$, we get:

$2I = \displaystyle \int\limits_{a}^{b} \dfrac{f(x)}{f(a + b - x) + f(x)} dx + \int\limits_{a}^{b} \dfrac{f(a + b - x)}{f(x) + f(a + b - x)} dx$

$\Rightarrow 2I = \displaystyle \int\limits_{a}^{b} \left(\dfrac{f(x)}{f(a + b - x) + f(x)} + \dfrac{f(a + b - x)}{f(x) + f(a + b - x)}\right) dx$

$\Rightarrow 2I = \displaystyle \int\limits_{a}^{b} \dfrac{\cancel{f(x) + f(a + b - x)}}{\cancel{f(a + b - x) + f(x)}} dx$

$\Rightarrow 2I = \displaystyle \int\limits_{a}^{b} dx$

$\Rightarrow 2I = x{\Large\vert}_a^b = b - a$

$\Rightarrow I = \dfrac{b - a}{2}$

Introduction To Integration