How do you integrate $int (3-2x)/(x(x²+3))$ using partial fractions?

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Answer 1 · 2017-12-19T19:21:47

$Lnx-1/2Ln(x^2+3)-(2sqrt3)/3*arctan(x/sqrt3)+C$

$int (3-2x)/[x*(x^2+3)*dx]$

I decomposed integrand into basic fractions

$(3-2x)/[x*(x^2+3)]$

= $A/x+(Bx+C)/(x^2+3)$

After expanding denominator,

$A*(x^2+3)+(Bx+C)*x=3-2x$

$(A+B)*x^2+Cx+3A=3-2x$

After equating coefficients, $A+B=0$ , $C=-2$ and $3A=3$

From them, $A=1$ , $B=-1$ and $C=-2$

Thus,

$int (3-2x)/[x*(x^2+3)*dx]$

= $int (dx)/x$ - $int ((x+2)*dx)/(x^2+3)$

= $Lnx-int x/(x^2+3)*dx-int 2/(x^2+3)*dx$

= $Lnx-1/2Ln(x^2+3)-(2sqrt3)/3*arctan(x/sqrt3)+C$

How do you integrate int (3-2x)/(x(x²+3))int (3-2x)/(x(x²+3)) using partial fractions?