A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
Evaluate the integral:
Indefinite Integral ((x^3 + x^2 + 2x + 1) / (x^2 + 1)(x^2 + 2)) dx
anonymous
 5 years ago
Evaluate the integral: Indefinite Integral ((x^3 + x^2 + 2x + 1) / (x^2 + 1)(x^2 + 2)) dx

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dw:1322597588202:dw

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so far I have x^3(A+C) + x^2 (B+D) + x(2A +C) +(2B + D)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But I believe I'm going in the wrong direction

slaaibak
 5 years ago
Best ResponseYou've already chosen the best response.2\[{{x^3 + x^2 + 2x + 1}\over{(x^2 + 1)(x^2 + 2)} } = {{Ax + B}\over{x^2 + 1}} + {{Cx + D}\over{x^2 + 2}}\] By partial franctions. Now:

slaaibak
 5 years ago
Best ResponseYou've already chosen the best response.2\[x^3 + x^2 + 2x + 1 = (Ax + B)(x^2 + 2) + (Cx + D)(x^2 + 1)\] \[x^3 + x^2 + 2x + 1 = Ax^3 + 2Ax + Bx^2 + 2B + Cx^3 + Cx + Dx^2 + D\] \[x^3 + x^2 + 2x + 1 = (A + C)x^3 + (B + D)x^2 + (2A + C)x + (2B + D)\] A + C = 1 B + D = 1 2A + C = 2 2B + D = 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Alright...but the answer I saw was (1/2)ln (x^2 + 1) + (1 / square(2)) arctan(x/square(2)) + C

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It's a frustrating problem

slaaibak
 5 years ago
Best ResponseYou've already chosen the best response.2Lets take it further then: By solving A, B, C, D: A = 1 C=0 B=0 D=1

slaaibak
 5 years ago
Best ResponseYou've already chosen the best response.2\[{{x^3 + x^2 + 2x + 1}\over{(x^2 + 1)(x^2 + 2)}} = {x \over {x^2 + 1} } + {{1}\over{x^2 + 2}}\]

slaaibak
 5 years ago
Best ResponseYou've already chosen the best response.2Now the fun starts! Now we integrate each one of those

slaaibak
 5 years ago
Best ResponseYou've already chosen the best response.2Now to integrate the first one: \[\int\limits {x \over {x^2 + 1} } dx\] \[x^2 = u\] \[2x dx = du\] \[xdx ={1\over2}du\] \[{1\over2}\int\limits{1\over{u + 1}} du\] =\[{1\over2}\ln(u + 1) + C\] resubstituting u with x^2: \[{1\over2}\ln(x^2 + 1) + C\]

slaaibak
 5 years ago
Best ResponseYou've already chosen the best response.2Do you need help with the second part of the integral?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0please if you don't mind

slaaibak
 5 years ago
Best ResponseYou've already chosen the best response.2\[\int\limits{1\over{x^2 + 2}} dx\] now, by factoring out the 2: \[\int\limits {1\over{2({1\over2}x^2 + 1)}} dx\] \[{1\over2}\int\limits{1\over{({1\over2}x^2 + 1)}}dx\] Putting the 1/2 inside the bracket: \[{1\over2} \int\limits { {1}\over{ ({{x}\over{\sqrt2}})^2 + 1 } }dx\] \[{x \over {\sqrt 2} } = u\] \[dx = \sqrt2 du\] \[{1\over2}\int\limits {{\sqrt2}\over{u^2+1}} du\] \[{{\sqrt2}\over2}\int\limits {{du}\over{u^2+1}}\] = \[{\sqrt2 \over 2} \arctan u + c\] substituting back: \[{\sqrt2 \over 2} \arctan({x \over \sqrt2}) + C\] since \[{\sqrt2 \over 2 } = {1\over \sqrt2}\] \[{1\over \sqrt2} \arctan({x \over \sqrt2}) + C\]
Ask your own question
Sign UpFind more explanations on OpenStudy
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
 Engagement 19 Mad Hatter
 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.