A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

El_Arrow

  • one year ago

don't understand how it got that answer please help

  • This Question is Closed
  1. El_Arrow
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    1 Attachment
  2. El_Arrow
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i understand the completing the square part i dont understand where the tangent came from

  3. El_Arrow
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @freckles @jim_thompson5910 @Zarkon

  4. El_Arrow
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hello?

  5. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\int\limits_{}^{}\frac{1}{u^2+1} du =\arctan(u)+C \\ \text{ and you have } \frac{1}{4} \int\limits \frac{dx}{\frac{1}{4}(x+3)^2+1} =\frac{1}{4} \int\limits \frac{dx}{(\frac{1}{2}(x+3))^2+1} \\ \text{ Let } u=\frac{1}{2}(x+3) \\ du=\frac{1}{2} dx\]

  6. jim_thompson5910
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    One way is to use a table of integrals (if you can recognize the form) http://integral-table.com/ look at equation (9) under the "Integrals of Rational Functions" section

    1 Attachment
  7. yolomcswagginsggg
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ...................__ ............./´¯/'...'/´¯¯`·¸ ........../'/.../..../......./¨¯\ ........('(...´...´.... ¯~/'...') .........\.................'...../ ..........''...\.......... _.·´ ............\..............( BROFIST ...........

  8. El_Arrow
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @freckles how did you get the 1/2 out front of (x+3)^2?

  9. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[a+b=b(\frac{a}{b}+1) \\ \text{ so like here we had } \\ (x+3)^2+4 =4[\frac{1}{4}(x+3)^2+1]\]

  10. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I just factored a 4 out on the bottom

  11. El_Arrow
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so (x+3)^2 is a and the 4 is b correct?

  12. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[(x+3)^2+4 \\ = \\ \frac{4}{4}(x+3)^2+4 \\ =4 \cdot \frac{1}{4}(x+3)^2+4 \cdot 1 \\ = \\ =4[\frac{1}{4}(x+3)^2+1]\]

  13. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yeah in that example yes you can say that which is perfectly fine

  14. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I put a few more lines in between there maybe it to make it more clear hopefully

  15. El_Arrow
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    okay thanks @freckles

  16. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    np

  17. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.