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Silenthill
Group Title
Integrate using partial fractions (x1)/(x^2(x^2+1))dx
 2 years ago
 2 years ago
Silenthill Group Title
Integrate using partial fractions (x1)/(x^2(x^2+1))dx
 2 years ago
 2 years ago

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eliassaab Group TitleBest ResponseYou've already chosen the best response.2
\[ \frac{x1}{x^2 \left(x^2+1\right)}=\frac{A}{x^2}+\frac{B}{x}+\frac{C x+D}{x^2+1} \]
 2 years ago

nbouscal Group TitleBest ResponseYou've already chosen the best response.0
\[ A(x^2+1)+Bx(x^2+1)+(Cx+D)x^2=x1 \]
 2 years ago

eliassaab Group TitleBest ResponseYou've already chosen the best response.2
Multiply bith sides by x^2 and make x =0, you get A= 1
 2 years ago

nbouscal Group TitleBest ResponseYou've already chosen the best response.0
Do you know how to take it from there, @Silenthill ?
 2 years ago

Silenthill Group TitleBest ResponseYou've already chosen the best response.0
yes thank you!
 2 years ago

eliassaab Group TitleBest ResponseYou've already chosen the best response.2
\[ \frac{x1}{x^2 \left(x^2+1\right)}=\frac{A}{x^2}+\frac{B}{x}+\frac{C x+D}{x^2+1} \] Multilpy both sides by x^2 + 1 and make x = i\[ \frac {i1} {1} = Ci + D= i+1\\ C=1\\ D=1\\ \]
 2 years ago

eliassaab Group TitleBest ResponseYou've already chosen the best response.2
\[ \frac{x1}{x^2 \left(x^2+1\right)}=\frac{A}{x^2}+\frac{B}{x}+\frac{C x+D}{x^2+1} \] Multiply both sides by x and let x goes to Infinity, you get 0 = B + C B=C=1 Putting everything together, you get \[ \frac{x1}{x^2 \left(x^2+1\right)}=\frac{1x}{x^2+1}\frac{1}{x^2}+\frac{1}{x} \]
 2 years ago

Silenthill Group TitleBest ResponseYou've already chosen the best response.0
thank you sir
 2 years ago

matricked Group TitleBest ResponseYou've already chosen the best response.0
given question equals x/(x^2(x^2+1) 1/(x^2(x^2+1) first part can be done usin u=x^2 and then applying partial factor method whereas the second part can be seperated as1/(x^2)1/(x^2+1) and then both the parts can be integrated easily
 2 years ago
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