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Antiderivatives

Primitive Functions
Given a function $f(x)$ defined on some interval $I$. Then a function $F(x)$ safisfies

$\displaystyle F^{\prime}(x) = f(x) $

for all $x \in I$ is called primitive function of $f(x)$.
Antiderivatives

Theorem 3..1   Let $F(x)$ be a primitive function of $f(x)$. Then every primitive function of $f(x)$ is given by $F(x) + c$, where $c$ is an arbitrary constant.

Proof Let $G(x)$ be a primitive function of $f(x)$. Then $G^{\prime}(x) = f(x)$ and since $F'(x) = f(x)$, $G^{\prime}(x) = F^{\prime}(x)$. Now let $H(x) = G(x) - F(x)$. Then $H'(x) = 0$. This means that $c = H(x) = G(x) - F(x)$, where $c$ is constant. Thus $G(x) = F(x) + c$. Note that $(F(x) + c)^{\prime} = F^{\prime}(x)$ $\ \blacksquare$

Antiderivatives
Every primitive function of $f(x)$ is called a antiderivative and denoted by $\int f(x)dx$.
The process of finding an antiderivative of $f(x)$ is calle antidifferentiation or integration.

Example 3..1   Integrate the following function

$\displaystyle \int (x^{-\frac{3}{2}} + \cos{x}) dx$

SOLUTION Note that $\displaystyle{(-2x^{-\frac{1}{2}})' = (-2)(-\frac{1}{2})x^{-\frac{3}{2}} = x^{-\frac{3}{2}}, (\sin{x})' = \cos{x}}$. Then since the derivative of a sum is the sum of the derivatives,

$\displaystyle (-2x^{-\frac{1}{2}} + \sin{x})' = x^{-\frac{3}{2}} + \cos{x}. $

$\displaystyle{\int (x^{-\frac{3}{2}} + \cos{x})\:dx = -2x^{-\frac{1}{2}} + \sin{x} + c \ \ensuremath{\ \blacksquare}}$

Exercise 3..1   Integrate the following function

$\displaystyle \int (\frac{\sin{x}}{\cos^{2}{x}} + \frac{1}{x}) dx$

SOLUTION Note that

$\displaystyle (\frac{1}{\cos{x}})' = \frac{\sin{x}}{\cos^{2}{x}}, \ (\log{x})' = \frac{1}{x}. $

Then since the derivative of a sum is the sum of the derivatives,

$\displaystyle (\frac{1}{\cos{x}} + \log{x})' = \frac{\sin{x}}{\cos^{2}{x}} + \frac{1}{x}. $

Thus, $\displaystyle{\int (\frac{\sin{x}}{\cos^{2}{x}} + \frac{1}{x})\:dx = \frac{\sin{x}}{\cos^{2}{x}} + \log{x} + c \ \ensuremath{\ \blacksquare}}$
Integration Formulas

Theorem 3..2   $\displaystyle{1. \ \int x^{\alpha} dx = \frac{x^{\alpha + 1}}{\alpha + 1} + c \hskip 0.3cm {\rm where} \ \alpha \neq -1}$

$\displaystyle{2. \ \int \frac{1}{x} dx = \log{\vert x\vert} + c}$

$\displaystyle{3. \ \int e^{x} dx = e^{x} + c ,\ \int a^x dx = \frac{a^x}{\log{a}} + c}$where $a > 0$

$\displaystyle{4. \ \int \sin{x}dx = -\cos{x} + c}$

$\displaystyle{5. \ \int \cos{x} dx = \sin{x} + c }$

$\displaystyle{6. \ \int \tan{x} dx = \log{\vert\sec{x}\vert} + c} = -\log\vert\cos{x}\vert + c$

$\displaystyle{7. \ \int \frac{1}{\cos^{2}{x}} = \tan{x} + c}$

$\displaystyle{8. \ \int \frac{1}{\sin^{2}{x}} = -\cot{x} +c}$

$\displaystyle{9. \ \int \frac{dx}{x^{2}-a^{2}} = \frac{1}{2a}\log{\vert\frac{x-a}{x+a}\vert} + c}$

$\displaystyle{10. \ \int \frac{dx}{x^{2}+a^{2}} = \frac{1}{a}\tan^{-1}{\frac{x}{a}} + c}$where $a \neq 0$

$\displaystyle{11. \ \int \frac{dx}{\sqrt{a^{2}-x^{2}}} = \sin^{-1}{\frac{x}{a}} + c }$where $\vert x\vert < a$

$\displaystyle{12. \ \int \frac{dx}{\sqrt{x^{2} + a}} = \log{\vert x + \sqrt{x^{2} + a}\vert} + c}$where $x^2 + a > 0$

NOTE Differentiate the right-hand side to get the integrand.
1. $\displaystyle{(\frac{x^{\alpha + 1}}{\alpha + 1} + c)^{\prime} = \frac{\alpha + 1}{\alpha + 1}x^{\alpha + 1 - 1} + c' = x^{\alpha}}$. Thus, $\displaystyle{\int x^{\alpha} dx = \frac{x^{\alpha + 1}}{\alpha + 1} + c}$
10.

$\displaystyle \left(\frac{1}{a}\tan^{-1}{\frac{x}{a}}\right)'$ $\displaystyle =$ $\displaystyle \frac{1}{a} \frac{1}{1 + \left(\frac{x}{a}\right)^2}\cdot \left(\... ...ght)' = \frac{1}{a^2}\cdot \frac{1}{1 + \frac{x^2}{a^2}} = \frac{1}{a^2 + x^2}.$  

Rules of Integration

Theorem 3..3   Suppose that integral of $f(x)$ and $g(x)$ exist. Then

$\displaystyle{1. \ \int\{f(x) \pm g(x)\}dx = \int f(x)dx \pm \int g(x)dx}$

$\displaystyle{2. \ \int kf(x)dx = k\int f(x)dx}\ \ {\rm where}\ $k $\ {\rm is \ constant}$

$\displaystyle{3. \ \int \frac{f'(x)}{f(x)}dx = \log\vert f(x)\vert + c}$

NOTE
$\displaystyle 1.\ \frac{d}{dx}(\int f(x)dx \pm \int g(x)dx)$ $\displaystyle =$ $\displaystyle \frac{d}{dx}\int f(x)dx \pm \frac{d}{dx}\int g(x)dx.$  
  $\displaystyle =$ $\displaystyle f(x) \pm g(x).$  

Thus, $\displaystyle{\int f(x)dx \pm \int g(x)dx = \int f(x)dx \pm \int g(x)dx}$
2. $\displaystyle{\frac{d}{dx}(k\int f(x)dx) = k \frac{d}{dx}(\int f(x)dx ) = k f(x)}$
3.

$\displaystyle \frac{d(\log\vert f(x)\vert)}{dx} = \frac{d(\log\vert u\vert)}{du}\frac{du}{dx} = \frac{1}{u}f'(x) = \frac{f'(x)}{f(x)}.$

Thus, ,

$\displaystyle \int \frac{f'(x)}{f(x)}\:dx = \log\vert f(x)\vert + c.$

Example 3..2   Integrate the following functions.
1. $\displaystyle{3\sin{x} + x^2}$ 2. $\displaystyle{\frac{1}{3 + x^2}}$3. $\displaystyle{\frac{1}{\sqrt{x^2 + 5}}}$

SOLUTION 1.
$\displaystyle \int(3\sin{x} + x^2)dx$ $\displaystyle \underbrace{=}_{{\rm Theorem}3.3-1}$ $\displaystyle \int 3\sin{x} dx + \int x^2 dx$  
  $\displaystyle \underbrace{=}_{{\rm Theorem}3.3-2}$ $\displaystyle 3\int \sin{x} dx + \int x^2 dx$  
  $\displaystyle \underbrace{=}_{{\rm Theorem}3.3-2}$ $\displaystyle -3\cos{x} + \frac{x^3}{3} + c\ensuremath{\ \blacksquare}$  

2.
$\displaystyle \int \frac{1}{3+x^2}\: dx$ $\displaystyle =$ $\displaystyle \int \frac{1}{(\sqrt{3})^2+x^2}\: dx$  
  $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{3}}\tan^{-1}\big(\frac{x}{\sqrt{3}}\big) + c\ensuremath{\ \blacksquare}$  

3.

$\displaystyle \int \frac{1}{\sqrt{x^2 + 5}}\: dx$ $\displaystyle =$ $\displaystyle \log\vert x + \sqrt{x^2 + 5}\vert + c\ensuremath{\ \blacksquare}$  

Exercise 3..2   Integrate the following functions.
1. $\displaystyle{\frac{x^2 - 1}{x^2 + 1}}$ 2. $\displaystyle{\frac{x^2 + 2x - 1}{x^2 + 1}}$ 3. $\displaystyle{\frac{1}{\sqrt{4 - t^2}}}$ 4. $\displaystyle{\frac{1}{x^2 - 4}}$

SOLUTION 1.

$\displaystyle \int \frac{x^2 - 1}{x^2 + 1}dx$ $\displaystyle =$ $\displaystyle \int \frac{x^2 + 1 - 2}{x^2 + 1}dx = \int (1 - \frac{2}{x^2 + 1})dx$  
  $\displaystyle =$ $\displaystyle x - 2\tan^{-1}{x} + c\ensuremath{\ \blacksquare}$  

2.
$\displaystyle \int \frac{x^2 + 2x - 1}{x^2 + 1}dx$ $\displaystyle =$ $\displaystyle \int \frac{x^2 + 1 + 2x - 2}{x^2 + 1}dx$  
  $\displaystyle =$ $\displaystyle \int (1 + \frac{2x-2}{x^2 + 1})dx$  
  $\displaystyle =$ $\displaystyle x + \int \frac{2x}{x^2+1}dx + \int\frac{-2}{x^2 + 1}dx$  
  $\displaystyle =$ $\displaystyle x + \log(x^2 + 1) - 2\tan^{-1}{x} + c\ensuremath{\ \blacksquare}$  

3.
$\displaystyle \int \frac{1}{\sqrt{4 - t^2}}\: dt$ $\displaystyle =$ $\displaystyle \int \frac{1}{\sqrt{2^2 - t^2}}\: dt = \sin^{-1}\big(\frac{t}{2}\big) + c\ensuremath{\ \blacksquare}$  

4.
$\displaystyle \int \frac{1}{x^2 - 4}\: dx$ $\displaystyle =$ $\displaystyle \int \frac{1}{(x-2)(x+2)}\: dx = \int \frac{1}{2}\left(\frac{1}{x-2} - \frac{1}{x+2}\right)\: dx$  
  $\displaystyle =$ $\displaystyle \frac{1}{2}\int \left(\frac{1}{x-2} - \frac{1}{x+2}\right)\:dx$  
  $\displaystyle =$ $\displaystyle \frac{1}{2}\left(\log\vert x-2\vert - \log\vert x+2\vert\right) + c = \frac{1}{2}\log\vert\frac{x-2}{x+2}\vert + c\ensuremath{\ \blacksquare}$  

Alternative solution By the rule of the integration, we have

$\displaystyle \int{\frac{1}{x^2 - 4}\:dx} = \int \frac{1}{x^2 - 2^2}\:dx = \log{\vert\frac{x-2}{x+2}\vert} + c$

Exercise A


1.
Work out the following integrals and check your answer by differentiation?D

(a) $\displaystyle{\int (2x-3) dx }$ (b) $\displaystyle{\int 5x^{4} dt}$ (c) $\displaystyle{\int \sqrt[3]{x^{2}} dx}$ (d) $\displaystyle{\int \frac{1}{\sqrt{x}} dx}$

(e) $\displaystyle{\int \frac{\sin{x}}{\cos{x}} dx}$ (f) $\displaystyle{\int \frac{1}{\cos^{2}x} dx}$ (g) $\displaystyle{\int \sin^{2}{x} dx}$ (h) $\displaystyle{\int \frac{1}{x^{2} - 4} dx}$

(i) $\displaystyle{\int \frac{1}{x^{2}+4}dx }$

Exercise B


1.
Work out the following integrals?D

(a) $\displaystyle{\int (3x^{-3} + 4x^{5}) dx }$ (b) $\displaystyle{\int (t^{3}+ \frac{1}{t^{2}+1}) dt}$ (c) $\displaystyle{\int -2\tan^{2}{x} dx}$

(d) $\displaystyle{\int \frac{x^{3} + 1}{\sqrt{x}} dx}$ (e) $\displaystyle{\int \frac{x^{2} + 5}{x^{2} + 4} dx}$ (f) $\displaystyle{\int \frac{1}{\sqrt{4 - t^{2}}} dt}$ (g) $\displaystyle{\int \cos^{2}{x} dx}$

(h) $\displaystyle{\int \frac{1}{\sqrt{t^{2} + 4}} dt}$ (i) $\displaystyle{\int \frac{1}{4 - x^{2}}dx }$



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