## LACHEEK989 2 years ago ∫1/x2+x+1

1. wio

Find the roots, then you can factor it. Then you want to try partial fraction decomposition.

2. LACHEEK989

I am lost at completing the square.

3. zepdrix

$\large x^2+bx$To complete the square, we take half of the $$\large b$$ term, and square it. $\large x^2+bx+\left(\frac{b}{2}\right)^2$Now we can't just add this number, it will change the value of our polynomial, we want to keep it balanced. So we have to also subtract this number. $\large \color{royalblue}{x^2+bx+\left(\frac{b}{2}\right)^2}-\left(\frac{b}{2}\right)^2$The reason for doing this is, the blue part will now factor down into a perfect square. A binomial containing x and half of the b term.$\large \color{royalblue}{\left(x+\frac{b}{2}\right)^2}-\left(\frac{b}{2}\right)^2$

4. zepdrix

Let's see if we can apply this to our problem here.

5. zepdrix

$\large x^2+x+1$The $$\large b$$ coefficient appears to be 1. (The middle term ~ because $$\large x$$ is the same as $$\large 1\cdot x$$ ). So to to complete the square, we'll take half of 1, and square it.$\large \color{orangered}{x^2+x+\left(\frac{1}{2}\right)^2}-\left(\frac{1}{2}\right)^2+1$I moved the 1 out of the way, we don't want to deal with that right now. Ok the orange part should give us a perfect square now,$\large \color{orangered}{\left(x+\frac{1}{2}\right)^2}-\left(\frac{1}{2}\right)^2+1$ Which will simplify further to,$\large \left(x+\frac{1}{2}\right)^2+\frac{3}{4}$

6. LACHEEK989

$4/3 \int\limits_{}^{}1/((4/3) u^2 +1) du$ What next?

7. LACHEEK989

(3/4)∫1/((4/3)u2+1)du I think this is right. What do I do next?

8. Outkast3r09

this seems much more easier than what is being done... but thats just me

9. Outkast3r09

$\frac{4}{3}\int\frac{1}{\frac{4}{3}u^2+1}du=\frac{\frac{4}{3}}{\frac{4}{3}} \int \frac{1}{u^2+1}du$

10. LACHEEK989

I agree. the book (answer in the back) and wolfram alpha I believe use a substitution here of 2/sqrt3 which gets complicated..

11. Outkast3r09

oh wait

12. Outkast3r09

is it $\frac{1}{4/3(u^2+1)}$

13. Outkast3r09

?

14. LACHEEK989

1/x2+x+1 tp begin with.

15. LACHEEK989

16. Outkast3r09

what is problem... like that is in the book

17. Outkast3r09

without anything done to it

18. LACHEEK989

$\int\limits_{}^{} (x^2 -x + 2) / (x^3 -1) \ \ dx$

19. LACHEEK989

I understand the partial fraction decomposition element to this. I am still lost at the point mentioned to complete the square. I haven't been taught yet.

20. Outkast3r09

alright well whered the partials you used?

21. Outkast3r09

I ask this because if you use partials you should've gotten a natural log for one of the fractions

22. wio

When you have $$u^2+1$$ it's a sign to try the trig sub $$u=\tan\theta$$.

23. LACHEEK989

yeah the answer so far is .. $2/3 \ln (x+1) + 1/6 \ln (x^2 + x + 1) -9/2\int\limits_{}^{} dx / (x^2 + x + 1)$

24. LACHEEK989

u = 4/sqrt3 ?

25. LACHEEK989

sorry 2/sqrt3 = u for the substitution?

26. wio

What is the original integral?

27. LACHEEK989

∫(x2−x+2)/(x3−1) dx

28. Outkast3r09

the last part you have to

29. Outkast3r09

complete the square and use trig sub

30. Outkast3r09

$x^2+x+1$ half of b = 1/2.. $\frac{b}{4}^2=(\frac{b}{2})^2=\frac{1}{4}$

31. Outkast3r09

if you add that you must subtract $\frac{1}{x^2+x+\frac{1}{4}+1-\frac{1}{4}}$

32. Outkast3r09

this is where he ot $\frac{1}{(x+\frac{1}{2})^2+\frac{3}{4}}$

33. LACHEEK989

then factor out the 3/4 in the bottom.

34. Outkast3r09

ok now you get to where you were before right?

35. LACHEEK989

right

36. JenniferSmart1

:'(

37. Outkast3r09

now let $x=\frac{a}{b}tan(\theta)$

38. LACHEEK989

is a/b 3/4 or does should I do a substitution like 2/sqrt 3 ?