Question

I need some help with checking this answer please:

\[\large\int\limits_4^5 \frac{x^3-3 x^2-9}{x^3-3 x^2} dx = x-\frac{3}{x}-\ln|3-x|+\ln|x|+C\]

I got this far using the partial fractions technique. I went to check my answer that I got on paper but Wolfram gives some craziness with hyperbolic tangents that I don't understand. So what should the answer be and wouldn't the natural log of 3-x be an issue because it gives non-real values?

The correct answer MUST have real values and be in terms of constants and logs. If somebody could show me what's going on here it would be great :-)

Answer 1

For my partial fractions I got:
\[x^3-3x^2-9=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x-3}\]
The A part turns into ln|x|, the B part turns into 3/x, the C part turns into -ln|3-x|

Answer 2

\[ \Large \int\frac{x^3-3x^2-9}{x^3-3x^2}\,dx=
\int\,dx-\int\frac{9}{x^2(x-3)}\,dx \]
so you have to solve
\[ \Large \frac{9}{x^2(x-3)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x-3} \]

Answer 3

Well, for one thing you have written the natural log of the absolute value of 3-x.

Answer 4

Yeah I used long division initially to pull out this:
\[1 + \frac{-9}{x^3-3x^2}\]

Answer 5

Integral of one with respect to x is just x, logically

Answer 6

I think I mistyped my second post, sry @helder_edwin

Answer 7

@Valpey oh yeah... so it's always a positive...

Answer 8

ln|3-(4)| = ln|-1| = ln 1 = 0

Answer 9

That's a nice simplification

Answer 10

/me waves hello @asnaseer :-) ty for stopping by!

Answer 11

you didn't need to do long division to simplify this in the first place. you could just have done this:\[\frac{x^3-3x^2-9}{x^3-3x^2}=\frac{x^3-3x^2}{x^3-3x^2}-\frac{9}{x^3-3x^2}=1-\frac{9}{x^3-3x^2}\]

Answer 12

secondly, the logs should not contain the absolute value symbols around them

Answer 13

Good point @asnaseer and why not? Without it wouldn't I have a non-real answer issue when I go to evaluate?

Answer 14

you should have done this:\[\int\limits_4^5 \frac{x^3-3 x^2-9}{x^3-3 x^2} dx =[ x-\frac{3}{x}-\ln(3-x)+\ln(x)]_4^5\]\[\qquad=[x-\frac{3}{x}+ln(\frac{x}{3-x})]_4^5\]\[\qquad=(5-\frac{3}{5}+ln(-\frac{5}{2}))-(4-\frac{3}{4}+ln(-\frac{4}{1}))\]\[\qquad=5-\frac{3}{5}-4+\frac{3}{4}+ln(-\frac{5}{2}\div-\frac{4}{1})\]

Answer 15

and the minuses cancel out inside the logs

Answer 16

*reading carefully*

Answer 17

BTW: waves hello back to @agentx5 :)

Answer 18

Oh... properties of logs...
\[\ln|a|-\ln|b|=\ln|\frac{a}{b}|\]

Answer 19

yes - but you shouldn't use the '|' symbols here

Answer 20

as they usually indicate absolute value

Answer 21

\[\qquad=5-\frac{3}{5}-4+\frac{3}{4}+ln(-\frac{5}{2}\div-\frac{4}{1}) \approx 3.45258\]

Disagrees with this check:

http://www.wolframalpha.com/input/?i=integrate+from+4+to+5+for+%28x^3-3x^2-9%29%2F%28x^3-3x^2%29
\(\approx \frac{17}{25}\)

Answer 22

I work it out to be exactly the same as the answer wolfram gives

Answer 23

I get:\[\frac{23}{20}+ln(\frac{5}{8})\approx 0.68\]

Answer 24

Ugh I see now I used the wrong base log on my calculator... /facepalm

Answer 25

Derp.

Answer 26

np :)

Answer 27

Thank you so much you were a great help! @TuringTest give that claymation-man a medal ^_^

Answer 28

you are more than welcome my friend! :)