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mukushla
 2 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
 2 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
 2 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
 2 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 \[\frac{ 1 }{ 2 } \int\limits_{}^{}\frac{ 1 }{ u }du = \frac{ 1 }{ 2 }\ln(u) = \frac{ 1 }{ 2 }\ln(x ^{2}+4)\] \[\left( \ln(23)  \frac{ 1 }{ 2 }\ln(4+4) \right)\left( \ln(03)  \frac{ 1 }{ 2 }\ln(0+4) \right) \] \[= \ln(1)\frac{ 1 }{ 2 }\ln(8)\ln(3)\frac{ 1 }{ 2 }\ln(4) = \ln(\frac{ 1 }{ 3\sqrt{2} })\]
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