Question

Help Needed Pls.

Answer 1

\[\int\limits_{0}^{1} \frac{ x^3-2x-7 }{ x^2-x-6 }\]

I started solving through long division and got \[\int\limits_{0}^{1} (x+1)+\frac{ 5x-1 }{ x^2-x-6 }\]

Then got stuck while integrating
\[\frac{ x^2 }{ 2} +x + \int\limits \frac{ 5x }{ x^2-x-6 }- \int\limits \frac{ 1 }{ x^2-x-6 }\]

Answer 2

Denominator factors, ya?
\(\large\rm x^2-x-6=(x-3)(x+2)\)
and then partial fraction decomposition.

Answer 3

Oh darn... I was hoping i didnt have to use PFD for this one.. lol
Oh whale...

Answer 4

but can \[\int\limits \frac{1 }{ x^2-x-6 }\] just be ln |x^2-x-6| ?

Answer 5

No, because the derivative of ln|x^2-x-6| is not 1/(x^2-x-6).
Chain rule, ya?

Answer 6

Its not? Eh.... lol

Answer 7

K so i got \[3 \ln |x-3|+2 \ln |x+2|\]
and then I do PFD for \[\int\limits \frac{ 1 }{ x^2-x-6 } \] as well?

Answer 8

PFD?

Answer 9

Partial Fraction Decomposition

Answer 10

~..~

Answer 11

:3

Answer 12

No no, do PFD from this point,\[\large\rm \int\limits \frac{5x-1}{(x-3)(x+2)}dx\]Don't break up the numerator.

Answer 13

Oh. shoot. Ooopsies

Answer 14

So I got \[\left[\frac{ x^2 }{ 2 }+x+\frac{ 14 }{ 3 } \ln |x-3|+\frac{ 11 }{ 2 }\ln|x+2| \right]| _{0}^{1}\]

Answer 15

After plugging in the values, I got...
\[\frac{ 3 }{ 2 } + \frac{ 14 }{ 3 }\ln2 +\frac{ 11 }{ 2 }\ln3 - \frac{ 14 }{ 3 }\ln3-\frac{ 11 }{ 2 }\ln2\] ....
Is this correct? D:

Answer 16

hmm how did you get 5 ?

Answer 17

5? where?

Answer 18

nvm lel..

Answer 19

xD

Answer 20

@zepdrix Did i do it correctly? :3

Answer 21

yea

Answer 22

Can someone approve of my answer? Idk if I did it correctly.

Answer 23

i got 11/5..hmm

Answer 24

Oh wait. It is over 5. I looked at it incorrectly.

Answer 25

But as your answer you got just 11/5?

Answer 26

\[\frac{ 3 }{ 2 } + \frac{ 14 }{ 5 } \ln2 +\frac{ 11 }{ 5 }\ln3-\frac{ 14 }{ 5 }\ln3-\frac{ 11 }{ 5 }\ln2\]

Answer 27

^Thats what i got...

Answer 28

looks good you can apply log rules to condense

Answer 29

@Nnesha you are not the only one that didn't know the abbreviation. they cut everything short these days.

Answer 30

haha true

Answer 31

Um... so would it be...
\[\frac{ 3 }{ 2 }+ \frac{ 14 }{ 11 }\ln2 -\frac{ 14 }{ 11 }\ln3\]?

Answer 32

Eh... Im lost

Answer 33

Nevermind, it accepted my answer without condensing. Thank you guys! Got it. Phew...

Answer 34

you should get 3/5 bec we have to add the coefficients hmmm \[\frac{ 14 }{ 5}\ln \left| 2 \right| +\frac{11}{5} \ln \left| 2 \right| = \color{blue}{\frac{3}{5} \ln \left| 2 \right|}\]
\[-\frac{14}{5} \ln \left| 3 \right| +\frac{11}{5} \ln \left| 2 \right| = \color{Red}{-\frac{3}{5} \ln \left| 3 \right|}\]
\[\frac{ 3 }{ 2 } \color{REd}{-\frac{3}{5} \ln \left| 3 \right|} +\color{blue}{\frac{3}{5} \ln \left| 2 \right|}\]
\[\frac{ 3 }{ 2 }-( \color{REd}{\frac{3}{5} \ln \left| 3 \right|} -\color{blue}{\frac{3}{5} \ln \left| 2 \right|})\]
\[\frac{ 3 }{ 2 }- \frac{3}{5} ( \color{REd}{\ln \left| 3 \right|} -\color{blue}{ \ln \left| 2 \right|})\]

division rule that's it :D
glad u dont have to do this :D

Answer 35

@Nnesha wouldnt the first one be\[\frac{ 25 }{ 5 }\ln2 = 5\ln2\]

Answer 36

Cos you were adding

Answer 37

no that supposed to be -11/5
my bad

Answer 38

Oh okay