Question

this is a irreducible Quadratic Factors question, its integral 0-1, dx/(x+1)(x^2+1)

Answer 1

you want to sole the integral, is that correctr?

Answer 2

\(\large\color{slate}{\displaystyle\int\limits_{0}^{1}\frac{1}{(x^2+1)(x+1)}~dx}\) like this?

Answer 3

i know that A=-1/2, but i am having a tough time with Bx+c

Answer 4

how did you set up the partial fraction expension ?

Answer 5

yes

Answer 6

1=A(x^2+1) + (Bx+c)(x-1)

Answer 7

so that when you multiplying 1st num * 2nd den. and vica verse you are going to be getting x^2 in both, and this way when you add them up (really subtracting, not adding) you will get 1.

Answer 8

\(\large\color{slate}{\displaystyle\frac{1}{(x^2+1)(x-1)}=\frac{Ax+B}{(x^2+1)}+\frac{C}{(x-1)}}\)

Answer 9

like that, now multiply every term times (x^2+1)(x-1)

Answer 10

so then it should be \[(Ax+B)(x-1)+C(x ^{2}+1)\]

Answer 11

=1

Answer 12

\(\large\color{slate}{\displaystyle 1=(Ax+B)(x-1)+C(x^2+1)}\)

Answer 13

now, remember this equation is true for any x-value, so plug in your x-values (of your choice)

Answer 14

ok so far so good

Answer 15

then i will use 1 first

Answer 16

yes, that appears to be it.

Answer 17

\(\large\color{slate}{\displaystyle 1=(Ax+B)(x-1)+C(x^2+1)}\)
\(\large\color{slate}{\displaystyle 1=(A(1)+B)(1-1)+C(1^2+1)}\)

Answer 18

C=?

Answer 19

1/2

Answer 20

yes, C=1/2.

Answer 21

Now, we have
\(\large\color{slate}{\displaystyle 1=(Ax+B)(x-1)+\frac{1}{2}(x^2+1)}\)

Answer 22

you can plug in other x-values

Answer 23

would 0 work

Answer 24

sure, why not.
(after plugging x=0, plug in another value, and then we would solve the system of equations for A and B)

Answer 25

so if i am correct it should be -1/2 =C

Answer 26

C=1/2 we already established that, didn't we?

Answer 27

-1/2 not 1/2

Answer 28

oh yes

Answer 29

clear why C=1/2 ?

Answer 30

so, we had:
\(\large\color{slate}{\displaystyle 1=(Ax+B)(x-1)+C(x^2+1)}\)

and then I plugged that value of C,
\(\large\color{slate}{\displaystyle 1=(Ax+B)(x-1)+\frac{1}{2}(x^2+1)}\)

Answer 31

is this clear so far?

Answer 32

yes, yes, i see. not thinking

Answer 33

so if we use zero, would use 0, would it be -1/2 = B

Answer 34

ok, now lets plug other x-values.

\(\large\color{slate}{\displaystyle 1=(Ax+B)(x-1)+\frac{1}{2}(x^2+1)\\[1.4em]}\)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
when x=0
\(\large\color{slate}{\displaystyle 1=(A(0)+B)(0-1)+\frac{1}{2}(0^2+1) \\[1.4em]}\)
\(\large\color{slate}{\displaystyle 1=(A(0)+B)(-1)+\frac{1}{2}\\[1.4em]}\)
\(\large\color{slate}{\displaystyle 1=-B+\frac{1}{2}\\[1.4em]}\)
\(\large\color{slate}{\displaystyle \frac{1}{2}=-B\\[1.4em]}\)
\(\large\color{slate}{\displaystyle -\frac{1}{2}=B\\[1.4em]}\)

Answer 35

\(\large\color{slate}{\displaystyle 1=(Ax+B)(x-1)+C(x^2+1)\\[1.4em]}\)
\(\large\color{slate}{\displaystyle 1=(Ax-\frac{1}{2})(x-1)+\frac{1}{2}(x^2+1)\\[1.4em]}\)

Answer 36

then solve for A.

Answer 37

I just plugged B and C

Answer 38

ok that is what i was about to say

Answer 39

very nice!

Answer 40

Alright, choose x=2 (I guess)

Answer 41

so then would A=1/4?

Answer 42

\(\large\color{slate}{\displaystyle 1=(Ax-\frac{1}{2})(x-1)+\frac{1}{2}(x^2+1)\\[1.4em]}\)
\(\large\color{slate}{\displaystyle 1=(A\color{blue}{(2)}-\frac{1}{2})(\color{blue}{(2)}-1)+\frac{1}{2}(\color{blue}{(2)}^2+1)\\[1.4em]}\)
\(\large\color{slate}{\displaystyle 1=2A-\frac{1}{2}+\frac{1}{2}(5)\\[1.4em]}\)
\(\large\color{slate}{\displaystyle 1=2A-\frac{1}{2}+\frac{5}{2}\\[1.4em]}\)
\(\large\color{slate}{\displaystyle 1=2A+2\\[1.4em]}\)
\(\large\color{slate}{\displaystyle -1=2A\\[1.4em]}\)
\(\large\color{slate}{\displaystyle A=-1/2\\[1.4em]}\)

Answer 43

now, lets substitute everything in.

Answer 44

wait wouldnt be 3/2?

Answer 45

why ?

Answer 46

I subtract 2 from both sides, not add.

Answer 47

tell me when you are done with 'A'

Answer 48

ok i see how A=-1/2

Answer 49

good

Answer 50

\(\large\color{slate}{\displaystyle \frac{1}{(x^2+1)(x-1)}=\frac{Ax+B}{(x^2+1)}+\frac{C}{(x-1)} \\[0.8em]}\)

we had C=1/2
A=B=-1/2

\(\large\color{slate}{\displaystyle \frac{1}{(x^2+1)(x-1)}=\frac{ -\frac{1}{2}x-\frac{1}{2}}{(x^2+1)}+\frac{\frac{1}{2}}{(x-1)\\[0.8em]}}\)

Answer 51

you can re-write this (multiplying top and bottom of both fractions times 2)

\(\large\color{slate}{\displaystyle \frac{1}{(x^2+1)(x-1)}=\frac{ -x-1}{2(x^2+1)}+\frac{1}{2(x-1)\\[0.8em]}}\)

Answer 52

or, alternatively

\(\large\color{slate}{\displaystyle \frac{1}{(x^2+1)(x-1)}=\frac{ -(x+1)}{2(x^2+1)}+\frac{1}{2(x-1)\\[0.8em]}}\)

Answer 53

Now, lets see what wolfram is getting.

Answer 54

This is the \(\rm\color{blueviolet}{\href{http:///www.wolframalpha.com/input/?i=%281%29%2F%28%28x-1%29%28x%5E2%2B1%29%29+partial+fractions}{the~~RESULT~~from~~wolfram}}\) .

Answer 55

ok i see how A=-1/2

Answer 56

now all i need to do is plug in the intervals for x and solve?

Answer 57

I substituted the A B and C for you above

Answer 58

but you need an integral of that from 0 to 1 though, correct ?

Answer 59

yes

Answer 60

\(\large\color{slate}{\displaystyle\int\limits_{0}^{1}\large\color{slate}{\frac{ -(x+1)}{2(x^2+1)}+\frac{1}{2(x-1)}}~dx}\)

Answer 61

like that ?

Answer 62

yes

Answer 63

so, the second part you can easily evaluate, correct ?

Answer 64

(the second fraction I mean)

Answer 65

yeah just put the 1 and 0 in for x

Answer 66

wel, you would first have to integrate that, and then plug in the limits (one and zero)

Answer 67

but that part is basic, now we will do \(\normalsize\color{slate}{\displaystyle\int\limits_{0}^{1}\large\color{slate}{\frac{ -(x+1)}{2(x^2+1)}}~dx}\) separately.

Answer 68

\(\large\color{slate}{\displaystyle\int\limits_{0}^{1}\large\color{slate}{\frac{ -x-1}{2(x^2+1)}}~dx}\)

\(\large\color{slate}{\displaystyle-\int\limits_{0}^{1}\large\color{slate}{\frac{ x}{2(x^2+1)}}~dx-\int\limits_{0}^{1}\large\color{slate}{\frac{ 1}{2(x^2+1)}}~dx}\)

Answer 69

then

\(\large\color{slate}{\displaystyle-\int\limits_{0}^{1}\large\color{slate}{\frac{ 2x}{4(x^2+1)}}~dx-\frac{1}{2}\int\limits_{0}^{1}\large\color{slate}{\frac{ 1}{(x^2+1)}}~dx}\)

Answer 70

see what I did so far ?

Answer 71

yes separated the integral, then took out a 1/2 to make it more simple

Answer 72

k, can you tell me what the second integral (general antiderivative) will be ?

Answer 73

let me write it fully actually

Answer 74

\(\large\color{slate}{\displaystyle-\int\limits_{0}^{1}\large\color{slate}{\frac{ x}{2(x^2+1)}}~dx-\int\limits_{0}^{1}\large\color{slate}{\frac{ 1}{2(x^2+1)}}~dx+\int\limits_{0}^{1}\large\color{slate}{\frac{ 1}{2(x-1)}}~dx}\)

Answer 75

we will go ahead an do the indefinite integral first for everything, and then plug in the limits.

Answer 76

are you ok with doing this?

Answer 77

well i see that you have to do the integration of the integral, which i think it is x/(x^2-x) but i dont think that is not right

Answer 78

no

Answer 79

k, lets go with the last integral.

Answer 80

what is the anti-derivatve of the 3rd integral ?

Answer 81

X/2x(x/2-x)

Answer 82

\(\large\color{slate}{\displaystyle-\int\limits_{0}^{1}\large\color{slate}{\frac{ x}{2(x^2+1)}}~dx-\int\limits_{0}^{1}\large\color{slate}{\frac{ 1}{2(x^2+1)}}~dx+\frac{ 1}{2}\rm{Ln}(x-1){{\huge |}_{ 0}^{1}}}\)

Answer 83

just like integral of 1/x is ln(x), so is integral of 1/(x-1) is ln(x-1)

Answer 84

then, the next one, is the inverse tangent function.

Answer 85

but we already got an improper integral

Answer 86

lets just focus on the last one for now, ok?

Answer 87

\(\large\color{blue}{\displaystyle\int\limits_{0}^{1}\frac{1}{2(x-1)}~dx}\)

Answer 88

ok

Answer 89

the way to do it, since it has a "1" boundary is:

\(\large\color{blue}{\displaystyle\lim_{~~~~~~a \rightarrow ~1^{{\LARGE \bf -}}}\left(\int\limits_{0}^{a}\frac{1}{2(x-1)}~dx\right)}\)

Answer 90

\(\large\color{blue}{\displaystyle\lim_{~~~~~~a \rightarrow ~1^{{\LARGE \bf -}}}\left[\frac{1}{2}\ln(x-1){{\huge |}_{ 0}^{a}}\right]}\)

Answer 91

well, the asolute value for ln

Answer 92

\(\large\color{blue}{\displaystyle\lim_{~~~~~~a \rightarrow ~1^{{\LARGE \bf -}}}\left[~~\frac{1}{2}\ln|x-1|~~{{\huge |}_{ 0}^{a}}\right]}\)

Answer 93

ok

Answer 94

what is the log of a very small number ?

Answer 95

for the second limit when x=0, the output is just zero, since ln(1)=0

so all you need is the first boundary

Answer 96

\(\large\color{blue}{\displaystyle\lim_{~~~~~~a \rightarrow ~1^{{\LARGE \bf -}}}\left[~~\frac{1}{2}\ln|a-1|~~\right]}\)

Answer 97

ok, i see that better now

Answer 98

when is ln(of a very small number ) ?