Original subtitles

1

In this video we're doing another partial fractions problem but we're talking about distinct quadratic

2

factors.

3

So we've been given this problem the integral of x square plus X plus 1 divided by quantity to X plus

4

one times quantity x squared plus 1 and we need to use partial fractions to evaluate this integral So

5

remember with partial fractions what we want to do is come up with a partial fractions decomposition

6

we want to simplify this fraction into its partial fractions decomposition so that we can integrate

7

that decomposition instead of this fraction which we can't integrate as is.

8

So how are we going to use partial fractions.

9

Well what we need to do first is recognize that we have distinct factors because we have a factor of

10

2 x plus 1 and a factor of x squared plus 1.

11

So they are distinct they're different from each other.

12

And then we have to recognize that 2 x plus 1 is a linear factor because this is X to the first power

13

and an X squared plus one is a quadratic factor because we have an x squared term.

14

So here's what our decompositions going to look like.

15

First of all we're going to take the original fraction we're going to put that on the left hand side

16

of this new equation that we're going to create.

17

So we're going to have 2 x plus one times quantity x squared plus 1.

18

OK so the original fraction exactly as is.

19

But on the right hand side when we have a linear factor we just put a single constant in the numerator

20

so we're going to do a divided by 2 x plus 1.

21

The single constant over the linear factor.

22

But then we're going to add to that here's where we put the quadratic factor X squared plus 1.

23

But whenever we have a quadratic factor we have to say B X plus C.

24

So we started with B because A was already taken.

25

If we had another linear factor here we would say plus D divided by the linear factor.

26

And if we had another quadratic factor we would say e x plus F divided by the other quadratic factor.

27

So that gives you an idea of how to set up a decomposition with linear factors and quadratic factors.

28

So now that we have this decomposition we need to simplify it and what we'll do is we'll multiply both

29

sides of this equation as always by the denominator from the left hand side so to X plus 1 times x squared

30

plus 1.

31

When we do that will multiply both of these factors by the left hand side.

32

We're going to get this entire denominator to cancel because this 2 x plus 1 will cancel with this 2

33

x plus 1 this x squared plus 1 will cancel with this x squared plus 1.

34

Leaving us with just x squared plus X plus 1 the numerator from the left hand side.

35

Then over here we multiply both of these factors by a divided by 2 x plus 1 will get the 2 x plus 1

36

to cancel with this 2 x plus 1.

37

Leaving us with just a times x squared plus 1.

38

The other factor then is here when we multiply 2 x plus 1 Times Square plus 1 by this b x plus C divided

39

by x squared plus 1 will get this x squared plus 1 to cancel with this x squared plus 1.

40

Leaving us with just the factor of 2 x plus 1 so will get plus B X plus C is important parentheses around

41

that multiplied by 2 x plus 1.

42

Now what we want to do is simplify the right hand side which will do by multiplying everything out.

43

So we'll distribute this across the x square plus 1 and we'll get a x squared plus a will foil out the

44

b x plus C times quantity 2 x plus 1 and will get X times 2 x is to be x squared.

45

We'll take B X times 1 and get B X. we'll take C times 2 x and get to see X and we'll take C times 1

46

and get c.

47

Now what we want to do is collect like terms.

48

So we're going to take all of our x squared terms together we'll say a x squared plus two B x squared.

49

That takes care of these two.

50

Then we'll take all of our x terms together so we'll say let's go ahead and put parentheses around these.

51

So are x terms we're going to have B X plus to see X that takes care of these two and then all grouped

52

together our constants or constants are going to be a plus C taking care of our last two terms then

53

what we want to do is factor out the x variable from each of these sets Apprentice's here so here we're

54

going to factor out an x squared.

55

So we're going to say a plus to B will be the only thing left when we factor in x squared out of X squared

56

plus 2 be x squared.

57

We pulled out the x squared.

58

We were just left with the A and A to B here in the second set of parentheses we're going to pull out

59

an x.

60

Leaving us with only B plus 2 c multiplied by X and then we'll leave our constants as they are.

61

So now what we want to do is recognize that over here on the left hand side we have 1 x squared plus

62

1 x plus 1.

63

What we want to do is equate coefficients from the left and the right hand side.

64

So we can say that the coefficient on x squared on the left hand side is one the coefficient on x square

65

on the right hand side is a plus to B we can say that the coefficient on X is 1 and the coefficient

66

on X is B plus 2 c.

67

And then finally that over here the constant is 1 and the constant over here is a plus C.

68

So now with that in mind what we can do is we can write equations where we set these things equal to

69

one another.

70

So we'll go ahead and say a plus two B is equal to 1 being plus 2 C is equal to 1 and a plus C is equal

71

to 1.

72

Now with this in mind let's go ahead and eliminate one of the variables.

73

We have three variables.

74

Let's go ahead and eliminate a.

75

So this last equation here a plus C is equal to 1.

76

We'll go ahead and subtract C from both sides and let's say a is equal to 1 minus C since a is equal

77

to one minus C we can plug one minus C in for a into this first equation.

78

So instead of a rate here we're going to get 1 minus C plus 2 B is equal to 1.

79

If we rewrite this we'll keep that to be.

80

We keep our minus C's all say to B minus C and then subtract 1 from both sides and we'll get to B minus

81

C is equal to zero.

82

Then we'll take our second equation here and put it with it so we'll say B plus 2 C is equal to 1.

83

Now what I want to do here is multiply through this second equation the left and the right hand side

84

all the terms by 2.

85

The reason is because if I multiply everything by two I'm going to get to B and then I can subtract

86

one equation from the other.

87

I'll get 2 B minus 2 B which is zero.

88

In other words I'll get my B's to cancel so I'll be able to solve for C.

89

So if I multiply through by 2 instead of B I'll get to B instead of two c I'll get 4 C.

90

And instead of one here I'll get two.

91

Now what I can do is subtract this bottom equation from the top equation and the result is going to

92

be to B minus two B which is zero so those cancel negative C minus 4 C's going to give me a negative

93

5 C and zero minus 2 is going to give me a negative 2.

94

If I divide both sides by negative 5 I get C is equal to negative 2 divided by negative 5.

95

The negative signs cancel and I get a positive two fifths.

96

Now I need to use this value of c to find values for a and b.

97

I already know a is equal to 1 minus C since C is two fifths.

98

I can say is equal to 1 minus two fifths.

99

I can call 1 here instead of 1 or call it 5.

100

So I have a common denominator and I'll get a is equal to 5 managed to is three or three fifths.

101

Now let's use this second equation here here again this one to find a value for B will go out and subtract

102

to see from both sides and we'll get B is equal to 1 minus 2 c plugging in two fifths for C.

103

I'm going to get B is equal to 1 minus two times two fifths is four fifths.

104

So I'll get four fifths.

105

I'll change my one to five deaths that I can have a common denominator and I'll get B is equal to 5

106

months for as 1 so I get 1 fifth.

107

So now I know that A is three fifths.

108

B is one fifth and C is two fifths.

109

With that in mind I can plug into my partial fraction decomposition and instead of a here I put three

110

fifths instead of B.

111

I'll put one fifth and instead of C I'll put two fifths.

112

And now this value here this entire right hand side is my partial fraction decomposition with the values

113

that I found for a b and c..

114

This is what I'm going to use to replace the original fraction and this will be a much easier integral

115

to evaluate.

116

So now instead of the original integral I can say the integral of and take this whole value here and

117

find that so I'm going to say three fifths divided by 2 x plus one plus one fifth X plus two fifths

118

divided by x squared plus 1 D x.

119

Now we can work on simplifying this integral.

120

So what we're going to do is we're going to split it into two separate integrals and we're going to

121

factor out the constants.

122

So we're going to bring this three fifths out in front and we're going to say three fifths times the

123

integral of one divided by 2 x plus 1 D X then I'm going to say plus here I'm going to factor out a

124

wonderful song and say one fifth time's the integral of x plus two.

125

Because when I multiply two by one fifth I get two fifths this value right here.

126

So I have to say X plus two divided by x squared plus 1 D x.

127

Now before I start evaluating integrals I also want to split the second integral into two integrals

128

of it's own.

129

So whenever you have multiple terms in the numerator of a fraction here you can split them into separate

130

fractions so instead of X plus two divided by x squared plus one I can say X divided by x squared plus

131

one plus and a separate fraction to divide it by x squared plus 1.

132

So if I rewrite this I'm going to say three fifths times the integral one over to X plus 1 D x plus

133

one fifth times the integral of x divided by x squared plus 1 x plus.

134

And I have to keep my one fifth here.

135

So plus one fifth time's the integral of two divided by x squared plus 1 the X and this first integral.

136

I can easily evaluate using my formula here remember that the integral of 1 divided by X is going to

137

be the natural log of the absolute value of x.

138

In other words whatever is in the denominator there is going to go inside of my absolute value bars

139

in the argument of my natural log function.

140

So I'm just going to take this three fifths leave it there as a coefficients of three fifths and then

141

I'm going to take my denominator to X plus 1 and say natural log of the absolute value of 2 x plus 1.

142

And the only thing I have to be careful of is I always have to apply changeroom and divide it by the

143

derivative of the denominator.

144

So here if you just have x by itself the derivative of X is 1.

145

So you would divide this by 1.

146

But of course that doesn't change anything which is why you don't see it.

147

But here we have 2 x plus 1.

148

The derivative of 2 x plus 1 is 2.

149

So I have to divide this by two which is the same as multiplying by 1 1/2.

150

So that takes care of our first integral.

151

The third integral here the last integral.

152

Keep in mind that I can pull this two out in front so I can call this one and change this to a 2.

153

Now I have two fifths out in front.

154

We know that the integral of 1 divided by x squared plus 1 that's a very common function.

155

That's the inverse tangent function.

156

So I can say plus two fifths times arctan or the inverse tangent function of x.

157

When you have the exact form 1 divided by x squared plus 1 the integral of that is the inverse tangent

158

function of x.

159

So that takes care of our last integral there and then this middle integral here will evaluate using

160

u substitution.

161

So for now let's just go ahead and say plus one fifth times the integral of x divided by x squared plus

162

1 D x.

163

Now for that integral we'll say you is equal to x squared plus 1 the derivative of u d u is going to

164

be equal to 2 x and then we add the D X and then if we solve this for Dayaks we get x is equal to D

165

you divide it by 2 x.

166

So now if we go ahead and make our substitutions for that integral we have that coefficient of 1 fifth

167

times the integral will keep that x in the numerator so x divided by we know X squared plus 1 is you.

168

So we'll have you in the denominator and X is going to be D u over to X which we just found here and

169

then we'll go ahead and add our other two terms.

170

Now inside of the integral you can see we're going to get X and the numerator to cancel with X in the

171

denominator.

172

This to here we can pull out in front.

173

This is a 2 in the denominator.

174

So it comes and moves into the denominator here.

175

Multiply it by the five and we end up with one tenth times the integral of one over you.

176

D'you hear we're going to multiply this one half by three fifths and we're going to get plus three over

177

10 natural log absoute value of 2 x plus 1 plus two fifths arctan or inverse tangent function of x and

178

then here for our integral the integral of one over you will be natural log of the absolute value of

179

you and then we'll go ahead and add to that the other terms we know that you as equal to x squared plus

180

one.

181

So we'll say one tenth natural log of the absolute value of x squared plus one will add these other

182

terms so three tenths natural log of absolute value to X plus one plus two fifths times inverse tangent

183

function of x and then we have to add the constant c to account for that constant of integration.

184

Now you could leave your final answer just like this.

185

Or if you wanted to you could factor out a one tenth and if you wanted to factor out one tenth you would

186

pull the one tent out in front like this.

187

And then we need to make sure we pulled out of every turn.

188

So obviously it's going to come out of this natural log absolute value of x squared plus 1.

189

Then we're also going to pull it out of three over 10 the 10 is going to get pulled out here from the

190

denominator.

191

But the three and the numerator is going to remain.

192

So we have plus three natural log absolute value to X plus 1.

193

And then here pulling one tenth out of two fifths.

194

So how do we do that easily.

195

Well basically all we're saying is we're going to factor a one tenth out of the two fifths.

196

So what do we have to multiply by one tenth in order to get two fifths.

197

And if you don't know how to do that in your head here's what you do.

198

What do we have to multiply by one tenth to get two fifths.

199

And then you would just multiply both sides by 10 and you would get x is equal to 20 over 5 for x is

200

equal to 4.

201

So when we factor 110 out of two fifths what's left over is force we would say plus for inverse tangent

202

of X and then we would close our brackets and then say plus C and that should make sense to us this

203

4 because if we multiply four by one tenth we get four over 10 which is the same as two over five.

204

So this is your final answer with the one Tende factored out in front in order to simplify it.