How do you integrate #(x^2 + 1)/(x(x + 1)(x^2 + 2))# using partial fractions?

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Answer 1 · 2017-11-18T19:28:34

#int (x^2+1)/[x*(x+1)*(x^2+2)]*dx#

=#1/2*Lnx#+#sqrt2/6*arctan(x/sqrt2)#+#1/12*Ln(x^2+2)#-#2/3*Ln(x+1)+C#

I decomposed integrand into basic fractions,

#(x^2+1)/[x*(x+1)*(x^2+2)]#

=#A/x+B/(x+1)+(Cx+D)/(x^2+2)#

After expanding denominator,

#A(x+1)(x^2+2)+Bx(x^2+2)+(Cx+D)x(x+1)=x^2+1#

#A*(x^3+x^2+2x+2)+B*(x^3+2x)+(Cx+D)*(x^2+x)=x^2+1#

#(A+B+C)*x^3+(A+C+D)*x^2+(2A+2B+D)*x+2A=x^2+1#

After equating coefficients,

#A+B+C=0#, #A+C+D=1#, #2A+2B+D=0# and #2A=1#

From these equations, #A=1/2, B=-2/3, C=1/6 and D=1/3#

Thus,

#int (x^2+1)/[x*(x+1)*(x^2+2)]*dx#

=#1/2*int (dx)/x-2/3*int (dx)/(x+1)+1/6*int ((x+2)*dx)/(x^2+2)#

=#1/2*Lnx-2/3*Ln(x+1)+1/12*int ((2x+4)*dx)/(x^2+2)#

=#1/2*Lnx-2/3*Ln(x+1)+1/12*int (2x*dx)/(x^2+2)+1/3*int (dx)/(x^2+2)#

=#1/2*Lnx-2/3*Ln(x+1)+1/12*Ln(x^2+2)+sqrt2/6*int (sqrt2*dx)/(x^2+2)#

=#1/2*Lnx-2/3*Ln(x+1)+1/12*Ln(x^2+2)+sqrt2/6*arctan(x/sqrt2)+C#