agentx5 3 years ago 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 :-)

1. agentx5

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|

2. helder_edwin

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

3. Valpey

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

4. agentx5

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

5. agentx5

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

6. agentx5

I think I mistyped my second post, sry @helder_edwin

7. agentx5

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

8. agentx5

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

9. agentx5

That's a nice simplification

10. agentx5

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

11. asnaseer

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

12. asnaseer

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

13. agentx5

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

14. asnaseer

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})$

15. asnaseer

and the minuses cancel out inside the logs

16. agentx5

17. asnaseer

BTW: waves hello back to @agentx5 :)

18. agentx5

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

19. asnaseer

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

20. asnaseer

as they usually indicate absolute value

21. agentx5

$\qquad=5-\frac{3}{5}-4+\frac{3}{4}+ln(-\frac{5}{2}\div-\frac{4}{1}) \approx 3.45258$ 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 $$\approx \frac{17}{25}$$

22. asnaseer

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

23. asnaseer

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

24. agentx5

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

25. agentx5

Derp.

26. asnaseer

np :)

27. agentx5

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

28. asnaseer

you are more than welcome my friend! :)