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Examples showing how various functions can be integrated

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METHOD 1

$\int_{}^{}x^n\:dx\;=\;\frac{1}{n\:+\:1}\:x^{n\:+\:1}\;+\;C$

(1)

Except when n = -1 Then

$\int_{}^{}\frac{1}{x\:dx}\;=\;Ln\,x\;+\;C$

(2)

<h5>Example</h5>

$\int_{}^{}\frac{1}{\sqrt[]{{2x\:+\:3}}}\:dx\;=\:\int_{}^{}(2x\:+\:3)^{-\frac{1}{2}}\:dx\:$

(3)

$=\;\frac{1}{2}\;.\;2(2x\:+\:3)^\frac{1}{2}\;+\;C$

(4)

<hr/>

METHOD 2

The integration of a rational algebraic function whose denominator factorizes.

$\int_{}^{}\frac{x}{(x\:-\:1)(x\:-\:2}\:dx\;=\;\int_{}^{}\left( \frac{-\,1}{x\:-\:1}\:+\:\frac{2}{x\:-\:2} \right)\:dx$

(5)

$=\:-\:Ln(x\:-\:1)\:+\:2\,Ln(x\:-\:2)\;+\;C$

(6)

$=\;Ln\left[K\frac{(x\:-\:2)^2}{(x\:-\:1)} \right]\;+\;C$

(7)

<hr/>

METHOD 3

The integration of rational algebraic functions whose denominators do not factorize

$\int_{}^{}\frac{f^{'}(x)}{f(x)}\;=\;Ln\:f(x)\;+\;C$

(8)

$\int_{}^{}\frac{2\,x\:+\:3}{x^2\:+\:3\,x\:+\:7}\;dx\;=\;Ln(x^2\:+\:3x\:+\:7)\;+\;C$

(9)

Blank

$\int_{}^{}\frac{a}{x^2\:+\:a^2}\:dx\;=\;tan^{-1}\left (\frac{x}{a}\right )\;+\;C$

(10)

<h5>Example</h5>

$\int_{}^{}\frac{1}{x^2\:+\:8x\:+\:25}\,dx\;=\;\frac{1}{3}\int_{}^{}\frac{3}{(x\:+\:4)^2\:+\:3^2}}\:dx$

(11)

$=\;\frac{1}{3}\:tan^{-1}\left (\frac{x\:+\:4}{3}\right )\;+\;C$

(12)

<h5>General example</h5>

$\int_{}^{}\frac{4x\:+\:5}{x^2\:+\:2x\:+\:2}\:dx\;=\;\int_{}^{}\left(\frac{2(2x\:+\:2)}{x^2\:+\:2x\:+\:2} \:+\:\frac{1}{(x\:+\:1)^2\:+\:1}\right)$

(13)

$2\:Ln(x^2\:+\:2x\:+\:2)\:+\:tan^{-1}(x\:+\:1)\;+\;C$

(14)

Really the general answer to this method will be in the form of $Ln(f(x))$ and $tan^{-1}(f(x))$

<hr/>

METHOD 4

The integration of an irrational algebraic fraction of the kind

$\frac{ax\:+\:b}{\sqrt[]{px^2\:+\:qx\:+\:s}}\;\:\:\;where\;p\:\neq\:0$

(15)

<h5>Example 1</h5>

$\int_{}^{}\frac{1}{\sqrt[]{x^2\:+\:2x\:-\:3}}\;=\;\int_{}^{}\frac{1}{\sqrt[]{(x\:+\:1)^2\:-\:4}}}$

(16)

$=\:cosh^{-1}(\frac{x\:+\:1}{2})\;+\;C$

(17)

Other forms

$\int_{}^{}\frac{b}{(x\:+\:a)^2\:-\:d^2}\:dx\;=\;b\:cosh^{-1}\:\frac{x\:+\:a}{d}$

(18)

$\int_{}^{}\frac{b}{\sqrt[]{(x\:+\:a)^2\:+\:d^2}}\;=\;b\:sinh^{-1}\:\frac{x\:+\:a}{d}$

(19)

$\int_{}^{}\frac{b}{\sqrt[]{d^2\:-\:(x\:+\:a)^2}}\;=\;b\:sin^{-1}\:\frac{x\:+\:c}{d}$

(20)

<h5>Example 2</h5>

$\int_{}^{}\frac{x\:+\:1}{\sqrt[]{x^2\:+\:2x\:+\:3}}\:dx\:=\;\sqrt[]{x^2\:+\:2x\:+\:3}\;+\;C$

(21)

but

$\frac{d\:\sqrt[]{x^2\:+\:2x\:+\:3}}{dx}\;=\;\frac{1}{2}(x^2\:+\:2x\:+\:3)^{-\frac{1}{2}}\:\cdot\:(2x\:+\:2)$

(22)

$=\:\frac{\frac{1}{2}(2x\:+\:2)}{\sqrt[]{x^2\:+\:2x\:+\:3}}$

(23)

From which it can be seen that

$\frac{d\:\sqrt[]{f\:x}}{dx}\;=\;\frac{\frac{1}{2}f^{'}x}{\sqrt[]{fx}}$

(24)

<h5>Example 3</h5>

$\int_{}^{}\frac{3x\:-\:5}{x^2\:+\:4x\:+\:7}\;dx$

(25)

$\int_{}^{}\left(\frac{3\:.\:\frac{1}{2}(2x\:+\:4)}{\sqrt[]{x^2\:+\:4x\:+\:7}} \,-\:\frac{11}{\sqrt[]{(x^2\:+\:2)^2}\:+\:3}}\right)\:dx$

(26)

$=\;3\:\sqrt[]{x^2\:+\:4x\:+\:7}\:-\:11\:sinh^{-1}\frac{x\:+\:2}{\sqrt[]{3}}\;+\;C$

(27)

In general the answer to this type are in the form

$\sqrt[]{}\;+\;(sinh\:or\:cosh\:or\:sin)^{-1}$

(28)

The integral of

$\sqrt[]{\frac{x\:-\:1}{2x\:+\:1}}\;dx$

(29)

can be found by multiplying top and bottom by

$\sqrt[]{x\:-\:1}$

(30)

thus

$\int_{}^{}\sqrt[]{\frac{x\:-\:1}{2x\:+\:3}}\;=\;\int_{}^{}\frac{x\:-\:1}{\sqrt[]{2x^2\:+\:x\:-\:3}}$

(31)

$=\:\frac{1}{2}\sqrt[]{2x^2\:+\:x\:-\:3}\;-\;\frac{5}{4\;\sqrt[]{2}}cosh^{-1}\frac{4x\:+\:1}{5}\;+\;C$

(32)

<hr/>

METHOD 5A

In the integral

$\int_{}^{}\frac{Ln\:x}{x}\;dx$

(33)

let

$U\;=\;Ln\:x$

(34)

$\therefore\;\:\;\frac{dU}{dx}\;=\;\frac{1}{x}$

(35)

$and\;\;\;dU\;=\;\frac{1}{x}\:dx$

(36)

thus the original equation can now be rewritten as :-

$\int_{}^{}\frac{Ln\:x}{x}\:dx\;=\;\int_{}^{}\frac{U}{x}\:.\:x\:dU$

(37)

and

$\int_{}^{}U\:dU\;=\;\frac{1}{2}U^2\;+\;C$

(38)

$\therefore\:\;\;\int_{}^{}\frac{Ln\:x}{x}\:dx\;=\;\frac{1}{2}(Ln\:x)^2\;+\;C$

(39)

<h5>Example</h5> To find the integral of

$\frac{\sqrt[]{x^3\:+\:a^2}}{x}\:dx$

(40)

let U =

$\sqrt[]{x^3\:+\:a^2}$

(41)

$\therefore\:\:\;U^2\:=\:x^3\:+\:a^2\;\;\;and\:\;\;2U\:\frac{dU}{dx}\:=\:3x^2$

(42)

$\therefore\;\;\;\frac{2}{3}\,U\:dU\;=\;x^2\:dx$

(43)

The integral can now be written as :-

$\int_{}^{}\frac{U}{U^2\:-\:a^2}\:dU$

(44)

$=\;\frac{2}{3}\int_{}^{}\frac{U^2}{U^2\:-\:a^2}\;dU$

(45)

$=\;\frac{2}{3}\int_{}^{}\left(1\:+\:\frac{a^2}{(U\:-\:a)(U\:+\:a)} \right)dU$

(46)

$=\;\frac{2}{3}\int_{}^{}\left(1\:+\:\frac{\frac{1}{2}a}{U\:-\:a}\:-\:\frac{\frac{1}{2}a}{U\:+\:a} \right)\:dU$

(47)

$=\;\frac{2}{3}\left[U\:+\:\frac{1}{2}\: a\:Ln\:\frac{U\:-\:a}{U\:+\:a}\right]\;+\:C$

(48)

$=\:\frac{2}{3}\left[\sqrt[]{x^3\:+\:a^2}\:+\:\frac{1}{2}\:a\:Ln\:\frac{\sqrt[]{x^3\,+\:a^2}\:-\:a}{\sqrt[]{x^3\:+\:a^2}\:+\:a} \right]$

(49)

<hr/>

METHOD 5B

The integral of an irrational function containing

$\sqrt[n]{ax\:+\:b}$

(50)

substitute

$U\;=\;\sqrt[n]{ax\:+\:b}\;\;\;i.e.\;\;\;U^n\:=\:ax\:+\:b$

(51)

$\therefore\;\;\;\frac{n}{a}\:\;U^{(n\:-\:1)}\:dU\;=\;dx$

(52)

So the integral is now rational in $U\:dU$

<h5>Example</h5> Find the integral of

$\int_{}^{}\frac{1}{x\:+\:\sqrt[]{2x\:-\:1}}\;dx$

(53)

substitute

$U\:=\:\sqrt[]{2x\:-\:1}\;\;\;i.e.\;\;\;x\:=\:\frac{U^2\:+\:1}{2}$

(54)

$\therefore\:\;\;U\:dU\;=\;dx$

(55)

thus the integral can be written as:-

$\int_{}^{}\frac{1}{\frac{U^2\:+\:1}{2}\:+\:a}\:U\:dU\;=\;2\,\int_{}^{}\frac{U}{(U^2\:+\:1)}\:dU$

(56)

$=\;2\:\int_{}^{}\left(\frac{1}{U\:+\:1}\:-\:\frac{1}{(U\:+\:1)^2} \right)$

(57)

$=\;2\:Ln\:(U\:+\:1)\:+\:\frac{2}{U\:+\:1}\:+\:C$

(58)

$\therefore\;\;\;\int_{}^{}\frac{1}{x\,+\:\sqrt[]{2x\:-\:1}}\:dx\;=\;2\,Ln\,(\sqrt[]{2x\,-\:1}\:+\:1\:+\:\frac{2}{\sqrt[]{2x\:-\:1}}\:+\:1}\;+\:C$

(59)

<hr/>

METHOD 6

The integration of Trigonometrical functions

<strong>A "Easy"</strong>

$\int_{}^{}cos\:x\:dx\;=\;sin\:x\:+\:C$

(60)

$\int_{}^{}sin\:x\:dx\:=\;-\:cos\:x\;+\;C$

(61)

$\int_{}^{}tan\:x\:dx\;=\;Ln\:sec\:x\;+\;C$

(62)

$\int_{}^{}sec^2\:x\,dx\:=\;tan\:x\;+\;C$

(63)

$\int_{}^{}sin^4\:cos\:x\:dx\;=\;\frac{1}{5}\:sin^5\,x\;+\;C$

(64)

<strong>B Using Trigonometrical formula</strong>

<h5>Example 1</h5> To find the integral of

$(cos \:5x\:cos\:2x)\;dx$

(65)

$but\;\;\;cos\:A\:+\:B\;=\;2\:cos\frac{A\:+\:B}{2}\:cos\frac{A\,-\:B}{2}$

(66)

from which it can be shown that

$\int_{}^{}cos\,5x\:cos\,2x\:dx\;=\;\frac{1}{2}\int_{}^{}(cos\,7x\:+\:cos\,3x)\:dx$

(67)

$=\:\frac{1}{14}\,sin\,7x\:+\:\frac{1}{6}sin\,3x\;+\;C$

(68)

] <h5>Example 2</h5>

$\int_{}^{}cos^2\:x\:dx\;=\;\frac{1}{2}\:\int_{}^{}(2\:cos\,2x\:+\:1)$

(69)

$=\;\frac{sin\:2x}{4}\:+\:\frac{x}{2}\:+\:C$

(70)

<strong>C Any Trigonometrical formula</strong>

To integrate any trigonometrical function such as f(sin x cos x) dx

$put\;\:\;\;t\;=\;tan\:\frac{x}{2}$

(71)

$but\:\;\:\;tan\:x\:=\:\frac{2\:tan\,\frac{x}{2}}{(1\:+\:tan^2\:\frac{x}{2})}$

(72)

$=\;\frac{2t}{1\:-\:t^2}$

(73)

<h5>Example 1</h5>

$\int_{}^{}cosec\:x\:dx\;=\;\int_{}^{}\frac{1}{sin\,x}\:dx$

(74)

$=\;\int_{}^{}\frac{1}{\frac{2t}{1\:+\:t^2}}\;X\;\frac{2}{1\:+\:t^2}\;dt$

(75)

$=\;\int_{}^{}\frac{1}{t}\:dt\;=\;Ln\:t\;+\;C$

(76)

$\therefore\;\;\;\int_{}^{}cosec\:x\:dx\;=\;Ln\,tan\frac{x}{2}\;+\;C$

(77)

<h5>Example 2</h5>

$\int_{}^{}sec\:x\;dx\;=\:\int_{}^{}\frac{1}{cos\:x}\:dx$

(78)

using the same substitution as above

$=\:\int_{}^{}\frac{1\:+\:t^2}{1\:-\:t^2}\;.\;\frac{2}{1\:+\:t^2}\:dt$

(79)

$=\:\int_{}^{}\left(\frac{1}{1\:-\:t}\:+\:\frac{1}{1\:-\:t} \right)\:dt$

(80)

$=\:-\:Ln\,(1\:-\:t)\:+\:Ln\,(1\:+\:t)\:+\:C$

(81)

$=\:Ln\:\frac{1\:+\:t}{1\:-\:t}\;+\:C$

(82)

$\therefore\;\;\;\int_{}^{}sec\:x\;dx\:=$

(83)

$\:Ln\frac{1\:+\:tan\,\frac{x}{2}}{1\:-\:tan\frac{x}{2}}\;+\;C$

(84)

<hr/>

METHOD 7

The integration of any hyperbolic function.

<strong>A Easy</strong>

$\intsech^2\:\theta\;d\theta\;=\;tanh\:\theta\;+\;C$

(85)

<strong>B Formula</strong>

$\int\:cosh^2\:\theta\:d\theta$

(86)

$=\:\frac{1}{2}\int(1\:+\:cosh\:2\theta)\:d\theta$

(87)

$=\:\frac{1}{2}\theta\:+\:\frac{1}{4}\:sinh\:2\,\theta\;+\;C$

(88)

<strong>C Any hyperbolic</strong>

$\intf(sinh\,\theta\:cos\,\theta)\:d\theta$

(89)

$put\;\:\;\;\;\;U\;=\;e^\phi$

(90)

Then

$sinh\:\theta\;=\;\frac{U\:-\:\frac{1}{U}}{2}\:=\:\frac{U^2\:-\:1}{2U}$

(91)

$cosh\:\phi\:=\:\frac{U\:+\:\frac{1}{U}}{2}\:=\:\frac{u^2\:+\:1}{2U}$

(92)

<h5>Example</h5>

$\intsech\:\phi\:d\phi\:=\:\int\frac{2U}{1\:+\:U^2}\:.\:\frac{1}{U}\:dU$

(93)

$=\:2\:tan{_1}\:U\:+\:C\;\;\;=\;\;\;2\:tan^{-1}\,e^\phi\:+\:C$

(94)

<hr/>

METHOD 8

Trigonometrical substitution Integration of irrational equations containing

$\sqrt[]{ax^2\:+\:bx\:+\:C}$

(95)

<h5>Example 1</h5>

$\int_{}^{}\frac{1}{x^2\:\sqrt[]{1\:-\:x^2}}\:dx$

(96)

$put\;\;\;\;x\:=\:sin\:\theta\;\;\:and\;\therefore\;\;\;dx\:=\:cos\:\theta\:d\theta$

(97)

$thus\;\;\;integral\:=\:\int_{}^{}\frac{1}{sin^2\theta\:cos\,\theta}\:cos\theta\:d\theta$

(98)

$=\:\int_{}^{}cosec^2\:\theta\:d\theta$

(99)

$=\:cot\:\theta\:+\:C\;\;=\:-\:\frac{\sqrt[]{1\:-\:x^2}}{x}\:+\:C$

(100)

<h5>Example 2</h5> Find the integral of

$\sqrt[]{(x^2\:+\:a^2)}\;dx$

(101)

Let $x = a sinh u$

$\therefore\;\;\;\frac{dy}{dx}\:=\:a\:cosh\,u$

(102)

$\frac{dy}{du}\:=\:\frac{dy}{dx}\:X\:\frac{dx}{du}\:=\sqrt[]{(x^2\:+\:a^2)}\:X\:a\:cosh\,u$

(103)

$but\;\;\;\sqrt[]{(x^2\:+\:a^2}\:=\:\sqrt[]{a^2\,sinh^2\,u\:+\:a^2)}=\:\sqrt[]{a^2\:cosh^2\,u}$

(104)

$thus\;\;\;\frac{dy}{du}\:=\:a^2\:cosh^2\,u\:=\:\frac{1}{2}\,a^2(1\:+\:cosh\,2u)$

(105)

$\therefore\;\;\;y\:=\:\frac{1}{2}\:a^2\int_{}^{}(1\:+\:cosh\:2u)\:du$

(106)

$=\:\frac{1}{2}\:a^2\:(u\:+\:\frac{1}{2}\,sinh\,2u)\:=\frac{1}{2}\:a^2\,u\:+\:\frac{1}{2}\:(sinh\,u\:cosh\,u)$

(107)

$=\:\frac{1}{2}\:a^2\:sinh^{-1}\frac{x}{a}\:+\:x\:\sqrt[]{(a^2\:+\:x^2)}$

(108)

<hr/>

METHOD 9

Integration by parts

$\frac{d\,(uv)}{dx}\:=\:u\:\frac{dv}{dx}\:+\:v\:\frac{du}{dx}$

(109)

$\therefore\;\;\;uv\:=\:\int_{}^{}u\:\frac{dv}{dx}\:dx\:+\:\int_{}^{} v\:\frac{du}{dx}\:dx$

(110)

$\int_{}^{}u\:\frac{dv}{dx}\:dx\:=\:uv\:-\:\int_{}^{}v\:\frac{du}{dx}\:dx$

(111)

this can also be written as:-

$\int_{}^{}\:u\:(v)^{'}\:dx\:=\:u\,v\:-\:\int_{}^{}v\:(u)^{'}\:dx$

(112)

<h5>Example 1</h5>

$\int_{}^{}\:x\:cos\,x\:dx\:$

(113)

$=\:\int_{}^{}x\:(sin\:x)^'}\:dx\:=\:x\:sin\,x\:-\:\int_{}^{}\:sin\,(x\:.\:1)$

(114)

$=\:x\:sin\,x\:+\:cos\,x\:+\:C$

(115)

<h5>Example 2</h5>

$\int_{}^{}Ln\:x\:dx\;=\;\int_{}^{}Ln\:x\:X\:1\:dx$

(116)

$\therefore\;\;\;\int_{}^{}Ln\,x\:(x)^'}\:dx\:=\:x\:Ln\,x\:-\:\int_{}^{}\:x\:X\:\frac{1}{x}\:dx$

(117)

$=\:x\:Ln\,x\:-\:x\:+\:C$

(118)

Integration by parts twice to regain the original integration.

$\int\:e^{3x}\:cos\,2x\:dx\:=\:\int\:e^{3x}\:(\frac{1}{2}\:sin\,2x)^'\:dx$

(119)

$=\:\frac{1}{2}\,e^{3x}\:sin\,x\:-\:\frac{3}{2}\int\:sin\,2x\:\:e^{3x}\:\:dx$

(120)

$=\:\frac{1}{2}\:e^{3x}\:sin\,x\:-\:\frac{3}{2}\:\int\:e^{3x}\:-\: (\frac{1}{2}\:cos\,2x)^{'}\:dx$

(121)

$=\:\frac{1}{2}\:e^{3x}\:sin\,x\:-\,\frac{3}{2}\left[-\:\frac{1}{2} e^{3x}\:cos\:2x\:+\:\frac{1}{2}\int\:e{3x}\:cos\,2x\:dx \right]}$

(122)

$\therefore\:\;\;\int\:e^{3x}\:cos\,2x\:dx\:=\:\frac{2}{13}\:e^{3x}\: sin\,2x\:+\:\frac{3}{13}\:e^{3x}\:cos\,2x\:+\:C$

(123)

<hr/>

METHOD 10

A large number of expressions can only be integrated by the method of successive reductions. This consists of making the integral dependent on a simpler integral, then again reducing this to one simpler still until a known form is found.

<h5>Example</h5>

$\int\:x^3\:cos\,x\:dx\:=\:\int\:x^3\:(sin\,x)^{'}\:dx$

(124)

$=\:x^3\:sin\,x\:-\:3\:\int\:x^2\:sin\,x\:dx$

(125)

$=\:x^3\:sin\,x\:-\:3\:\int\:x^2\:(\:-\:cos\,x)^{'}\:dx$

(126)

$=\:x^3\:sin\,x\:+\:3\:x^2\:cos\,x\:-\:6\:\int\:x\:cos\,x\:dx$

(127)

$=\:x^3\:sin\,x\:+\:3\:x^2\:cos\,x\:-\:6\:\int\:x\:(sin\,x)^{'}\:dx$

(128)

$=\:x^3\:sin\,x\:+\:3\:x^2\:cos\,x\:-\:6\:x\:sin\,x\:+\:6\:\int\:sin\,x\:dx$

(129)

WELL THAT IS ABOUT THAT! THERE ARE ACTUALLY OTHER METHODS, BUT MOST OF THESE REQUIRE NUMERICAL INTEGRATION AND COMPUTER PROGRAMS. See simpson, gauss, trapezoidal

Methods of Integration

METHOD 1

METHOD 2

METHOD 3

METHOD 4

METHOD 5A

METHOD 5B

METHOD 6

METHOD 7

METHOD 8

METHOD 9

METHOD 10