How do you integrate #1/(x^3 +4x)# using partial fractions?

3 Answers
Nov 13, 2017

The answer is #=1/4ln(|x|)-1/8ln(x^2+4)+C#

Explanation:

Perform the decomposition into partial fractions

#1/(x^3+4x)=1/(x(x^2+4))=A/x+(Bx+C)/(x^2+4)#

#=(A(x^2+4)+x(Bx+C))/(x(x^2+4))#

The denominators are the same, compare the numerators

#1=A(x^2+4)+x(Bx+C)#

Let, #x=0#, #=>#, #1=4A#, #=>#, #A=1/4#

Coefficients of #x^2#

#0=A+B#, #=>#, #B=-A=-1/4#

Coefficients of #x#

#0=C#

Therefore,

#1/(x^3+4x)=(1/4)/x+((-1/4)x)/(x^2+4)#

So,

#int(dx)/(x^3+4x)=int(1/4dx)/x+int((-1/4dx)x)/(x^2+4)#

#=1/4int(dx)/x-1/4int(xdx)/(x^2+4)#

Perform the second integral by substitution

Let #u=x^2+4#, #=>#, #du=2xdx#

So

#1/4int(xdx)/(x^2+4)=1/8int(du)/u=1/8lnu=1/8ln(x^2+4)#

#int(dx)/(x^3+4x)=1/4ln(|x|)-1/8ln(x^2+4)+C#

Nov 13, 2017

#1/4*Lnx-1/8*Ln(x^2+4)+C=1/8*Ln(x^2/(x^2+4))+C#

Explanation:

#int dx/(x^3+4x)#

=#int dx/[x*(x^2+4)]#

=#1/4*int (4*dx)/[x*(x^2+4)]#

=#1/4*int [(x^2+4)*dx]/[x(x^2+4)]#-#1/4*int [x^2*dx]/[x(x^2+4)]#

=#1/4*int (dx)/x#-#1/4*int (x*dx)/(x^2+4)#

=#1/4*Lnx-1/8*Ln(x^2+4)+C#

=#1/8*Ln(x^2/(x^2+4))+C#

Nov 13, 2017

# 1/8ln|x^2/(x^2+4)|+C.#

Explanation:

Here is another way to solve the Problem, but without the

use of Partial Fractions.

Let, #I=int1/(x^3+4x)dx=int1/{x(x^2+4)}dx=intx/{x^2(x^2+4)}dx,# or,

#I=1/2int(2x)/{x^2(x^2+4)}dx.#

Subst. #x^2=t rArr 2xdx=dt.#

#:. I=1/2int1/{t(t+4)}dt=1/2int1/(t^2+4t)dt,#

#=1/2int1/{(t^2+4t+4-4)}dt,#

#=1/2int1/{(t+2)^2-2^2}dt.#

Since, #int1/(y^2-a^2)dy=1/(2a)ln|(y-a)/(y+a)|,#

#I=1/2*1/(2*2)ln|{(t+2)-2}/{(t+2)+2}|,#

#=1/8ln|t/(t+4)|,#

# rArr I=1/8ln|x^2/(x^2+4)|+C.#