Question

Challenge Part 1
find the area of the region enclosed by the x-axis and the curve y=ln(x^2-x+2) on [0,1]

Answer 1

Try using integration by parts and later on a partial fraction decomposition.

Answer 2

\[\large \int1\cdot \ln(x^2-x+2)=x\ln(x^2-x+1)-\int x\cdot \frac{1}{x^2-x+1}(2x-1) \]

Answer 3

\[\large x\ln(x^2-x+1)-\int\frac{2x^2-x}{x^2-x+1}dx \]
Not quite ready for a partial fraction decomposition, but after a longhand division things should look better.

Answer 4

i should have said, i know this but partial fractions is hard not integration by parts

Answer 5

are you sure it is a plus 2 not a minus?

Answer 6

heck sure...if its minus...its not a challenge anymore

Answer 7

if it was a minus, it would be easy

Answer 8

ya i know

Answer 9

did you try integration tables or are you not allowed them?

Answer 10

If not via partial fraction decomposition, then I can only guess that you want to use a trig identity.

Answer 11

i dont even know what integral tables are

Answer 12

they would solve this in an instant lol

Answer 13

haha then i guess we are not allowed to use them

Answer 14

ready?

Answer 15

this is only the \[\int{(x^2-x+2})dx\]
I am not carrying the \[2\pi\] around through the whole thing

Answer 16

So to start let's look at the polynomial and try to "simplify" it. We can complete the square to get it in the form \[\int{\ln{|(x-.5)^2+1.75|}}dx\]

Answer 17

from here let u=x-.5 then du=dx

Answer 18

then we have ∫ln|(u)^2+1.75|dx

Answer 19

now this is equivalent to \[\int\limits_{}^{} \ln[(u+\frac{\sqrt{7} }{ 2 })(u-\frac{\sqrt{7} }{ 2 })]dx \]

Answer 20

that should be a du not dx

Answer 21

now by ln rules and int rules we can make it ∫ln[(u+7√2)]du+∫ln[(u−7√2)]du

Answer 22

now, yet another substitution let to get ∫ln|v|dv+∫ln|g|dg

Answer 23

then plug u back in first then put your x's back

Answer 24

Careful, it seems to me that you're missing something at one step.

Answer 25

final answer that I got in terms of u was was \[\int ln (x^2-x+2) dx=(u+\frac { \sqrt{7}}{2})ln|(u+\frac { \sqrt{7}}{2})|-(u+\frac { \sqrt{7}}{2})+(u-\frac { \sqrt{7}}{2})ln|(u-\frac { \sqrt{7}}{2})+(u-\frac { \sqrt{7}}{2})+C\]

Answer 26

Yea, I accidentally dropped the ln at the start but I picked it back up

Answer 27

You said yourself that u^2+1.75, but you cannot factor this by (u-x)(u+x)=u^2-x^2

Answer 28

or do I misread that step?

Answer 29

(Sorry I am on my windows machine, it has problems with LaTeX at some point, so I might have confused it)

Answer 30

yea, I used conjugates where 1.75=((sqrt7)/2)^2

Answer 31

right or did I screw that step up? It is quite possible

Answer 32

I see that, but that would resunt in a minus still, not in a plus.

Answer 33

ahh it is a minus, alas

Answer 34

it's really a good idea though, I was having troubles with the plus sign before myself, I tried with getting it into the complex plane first, there it would be possible for this substitution, but the result remains imaginary.

Answer 35

yea, I did that first as well. so now at least it is just \[\int(ln|u^2+1.75|du)\]

Answer 36

hmm curious

Answer 37

if you take the squareroots for it and multiply them together, would trig sub be helpful?

Answer 38

I dont know why but I like to bring challenge question...so complicated and so interesting...

Answer 39

haha yea you do

Answer 40

the trig substitution looks way more promising than everything else I tried so far, but it remains complex, I can relog reload quick and then try to share what I got.

Answer 41

ooh wait why wolfram alpha this

Answer 42

why not*

Answer 43

So far this seems interesting
\[\Large \ln\left((x-\frac{1}{2})^2+\frac{7}{4}\right) \]

After carrying through with the trig substitution it looks like this:
\[\Large \ln\left(\left(\frac{\sqrt{7}}{2}\tan\alpha\right)^2 \right)\cdot \frac{\sqrt{7}}{2}\sec\alpha\tan\alpha \]

Answer 44

If I didn't make any careless mistakes.

Answer 45

setting \[\Large u=\frac{\sqrt{7}}{2} \sec\alpha\] to obtain the above.

Answer 46

but you still have a ln multiplication

Answer 47

optically it's pretty neat, especially with the exponent which could also be factored about, but it doesn't help with any substitution I can think of.

Answer 48

yes,
\[\Large 2\ln\left(\left(\frac{\sqrt{7}}{2}\tan\alpha\right) \right)\cdot \frac{\sqrt{7}}{2}\sec\alpha\tan\alpha d\alpha\]

Answer 49

that's as good as it gets for me, until now.

Answer 50

interesting, but i'm not sure how to do that trig sub. but I will def look at it when I have time again

Answer 51

:O

Answer 52

interesting stuff...i missed everything lol

Answer 53

So far it's an useless result anyway, no substitution will help with the above integral, I mean, the one I tried to derive. (-:

Answer 54

2.64575 tan^(-1)(0.755929 x)+x (-2.+log(1.75+x^2)) - This is wolfram answer

Answer 55

it is wayy tricky

Answer 56

I understand why there is an arc tan it it, that also speaks at least somehow for a trig substitution, maybe not quite the one I did though.

Answer 57

check this out

Answer 58

with parts,
\[\Large x\ln(x^2-x+2)-\int\frac{2x^2-x}{x^2-x+2}dx \\
\Large x\ln(x^2-x+2)-\int\left(\frac{x}{x^2-x+2}-\frac{4}{x^2-x+2}+2\right) \]

Answer 59

this seems promising, is it what they/you did?

Answer 60

hmm almost. the above could be integrated, will be arctan functions, just messy to deal with.

Answer 61

\[\Large \int \left(\frac{x}{(x-.5)^2+1.75}+4\frac{1}{(x-.5)^2+1.75}+2\right)dx \]

Answer 62

so whats the answer...

Answer 63

well do you agree with the by parts method so far?

Answer 64

ya

Answer 65

\[\large \int\ln(x^2-x+2)=x\ln(x^2-x+2)-\int\frac{2x^2-x}{x^2-x+2}dx \]
So from here on I will just talk about the very right integral, or more generally, the quotient :
\[\Large \frac{2x^2-x}{x^2-x+2}=\frac{2x^2}{x^2-x+2}-\frac{x}{x^2-x+2}\]
or after a long division
\[\Large \frac{2x^2-x}{x^2-x+2}=2+\frac{2x-4}{x^2-x+2}-\frac{x}{x^2-x+2}
\]

Answer 66

\[\Large \frac{2x^2-x}{x^2-x+2}=2+\frac{x}{x^2-x+2}-\frac{4}{x^2-x+2} \]

Answer 67

\[\Large \frac{2x^2-x}{x^2-x+2}=2+\frac{x}{(x+\frac{1}{2})^2+\frac{7}{4}}-\frac{4}{x+\frac{1}{2})^2+\frac{7}{4}} \]

Answer 68

These terms can be integrated.

Answer 69

The first one will just give a 2x, after that comes a logarithmic property (u sub) with an arctan function and the last one again is an arctan function, so if I didn't make any mistakes, this is in harmony with the result delivered by the wolf.

http://www.wolframalpha.com/input/?i=int+ln%28x%5E2-x%2B2%29

Answer 70

got it...i dont trust wolframalpha for calculus stuff...and check the three pages i posted before...are they correct...those r done by wolframalpha

Answer 71

Hmm don't quite remember all the steps, I would have to review, but most of it seemed clear to me. It's a good thing to mistrust the wolf in general. The steps by wolf seem a bit lengthy though.

Answer 72

ya i know

Answer 73

Do you concur with these steps @FibonacciChick666 ?

Answer 74

I think this is above my pay-grade, maybe when I grow up I could help lol.