A community for students.
Here's the question you clicked on:
 0 viewing
amorfide
 3 years ago
I need to know how to integrate this
amorfide
 3 years ago
I need to know how to integrate this

This Question is Closed

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.0use partial fractions here\[\frac{3x+4}{(x^2+4)(x3)}=\frac{Ax+B}{x^2+4}+\frac{C}{x3}\]and find A,B and C

carson889
 3 years ago
Best ResponseYou've already chosen the best response.2Split using partial fractions: \[\frac{ 3x+4 }{ (x ^{2}+4)(x3) } = \frac{ A }{ x3 } + \frac{ Bx + C}{ (x ^{2} + 4) } = \frac{ A(x^{2} +4)+(Bx+C)(x3) }{ (x ^{2}+4)(x3) }\] Set equal to 3, looking at just numerator: \[3*3 + 4 = 13 = A(9+4) + (3B+C)(0) = 13A \rightarrow A = 1\] Set x equal to 0: \[0 + 4 = 4 = 4A + C \rightarrow 4 = 4+C \rightarrow C = 0\] Set x equal to 1: \[3 + 4 =A(1+4) + (B+C)(13) = 5A 2B 2C \rightarrow 5  2B  0 \rightarrow 2 = 2B \rightarrow B = 1\] Therefore: \[\frac{ 3x+4 }{ (x ^{2}+4)(x3)} = \frac{ 1 }{ x3 }  \frac{ x }{ x ^{2} + 4 }\] Now integrate those.

amorfide
 3 years ago
Best ResponseYou've already chosen the best response.0thank you very much, you have no idea how much i love you xD

carson889
 3 years ago
Best ResponseYou've already chosen the best response.2No problemo The integral of 1/(x3) is ln(x3) and the integral of x/(x^2 + 4): usub: u = x^2 + 4, du = 2x dx \[\frac{ 1 }{ 2 } \int\limits_{}^{}\frac{ 1 }{ u }du = \frac{ 1 }{ 2 }\ln(u) = \frac{ 1 }{ 2 }\ln(x ^{2}+4)\] \[\left( \ln(23)  \frac{ 1 }{ 2 }\ln(4+4) \right)\left( \ln(03)  \frac{ 1 }{ 2 }\ln(0+4) \right) \] \[= \ln(1)\frac{ 1 }{ 2 }\ln(8)\ln(3)\frac{ 1 }{ 2 }\ln(4) = \ln(\frac{ 1 }{ 3\sqrt{2} })\]
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.