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#