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

Question

Paisley Johnson · Accepted Answer

#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#

Emilia Aguilar · Answer

undefined To integrate the function (x^2 + 1)/(x(x + 1)(x^2 + 2)) using partial fractions, first, express the rational function as the sum of partial fractions. After simplification, the partial fraction decomposition will have the form:

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

Next, find the values of A, B, C, and D by equating the numerators of the partial fractions to the original function. After finding the values of A, B, C, and D, integrate each partial fraction separately.

The final step is to combine the integrals of the partial fractions to obtain the integral of the original function.