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Integrate 2x+3/((x-2)(x^2+3))
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gonna have to do partial fractions i think, this may take a while, but i'll try
Thank you I'm taking an exam right now please help
2x+3/((x-2)(x^2+3))=A/(x-2)+(Bx+C)/(x^2+3) 2x+3=A(x^2+3)+(Bx+C)(x-2) =Ax^2+3A+Bx^2-2Bx+Cx-2C =(A+B)x^2+(-2Bx+C)x+(3A-2C)=2x+3 setting coefficients equal: A +B =0 -2B+C=2 3A -2C=3 Solve the system to get A=1, B=-1, C=0 so now our integral is, just wait...
\[\int\limits1/(x+2)-x/(x^2+3)dx=\ln (x+2)-(1/2)\ln (x^2+3)+C\]
actually it should have absolute value signs: \[\ln \left| x+2 \right|-(1/2)\ln \left| x^2+3 \right|+C\]
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