#int(4x^2-x+12)/(x^3+4x)dx#?

2 Answers
Feb 20, 2018

# 3ln|x|+1/2ln(x^2+4)-1/2arc tan(x/2)+C, or, #

# ln|x^3*sqrt(x^2+4)|-1/2arc tan(x/2)+C#.

Explanation:

Let, #I=int(4x^2-x+12)/(x^3+4x)dx=int(4x^2-x+12)/{x(x^2+4)}dx#.

We will use Heaviside's Method to decompose the

integrand #(4x^2-x+12)/{x(x^2+4)}# into Partial Fraction :

#color(red)(4x^2-x+12)/{color(blue)(x)(color(red)(x^2+4))}=A/x+(Bx+C)/(x^2+4)," say for some "A,B,C in RR#.

Then, we have, #A=[color(red)((4x^2-x+12)/(x^2+4))]_(color(blue)(x=0))=12/4=3#.

Sub.ing #A=3# above, we have,

#(4x^2-x+12)/{x(x^2+4)}-3/x=(Bx+C)/{x(x^2+4)}, i.e.,#

#(Bx+C)/{x(x^2+4)}={(4x^2-x+12)-3(x^2+4)}/{x(x^2+4)}#,

#=(x^2-x)/{x(x^2+4)}#.

#rArr (Bx+C)/(x^2+4)=(x-1)/(x^2+4)," giving, "B=1, C=-1#.

Altogether, #(4x^2-x+12)/{x(x^2+4)}=3/x+(x-1)/(x^2+4)#.

#:. I=int3/xdx+int(x-1)/(x^2+4)dx#,

#=3ln|x|+intx/(x^2+4)dx-int1/(x^2+2^2)dx#,

#=3ln|x|+1/2int(2x)/(x^2+4)dx-1/2arc tan(x/2)#,

#=3ln|x|+1/2int{d/dx(x^2+4)}/(x^2+4)dx-1/2arc tan(x/2)#.

#rArr I=3ln|x|+1/2ln(x^2+4)-1/2arc tan(x/2)+C, or, #

#I=ln|x^3*sqrt(x^2+4)|-1/2arc tan(x/2)+C#.

Feel & Spread the Joy of Maths.!

Feb 20, 2018

The answer is #=3ln(|x|)+1/2ln(x^2+4)-1/2arctan (x/2)+C#

Explanation:

The degree of the denominator is #># than the degree of the numerator

The denominator is

#=(x^3+4x)=(x(x^2+4))#

Perform the decomposition into partial fractions

#(4x^2-x+12)/(x^3+4x)=(4x^2-x+12)/(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

#4x^2-x+12=A(x^2+4)+x(Bx+C)#

Let #x=0#, #=>#, #12=4A#, #=>#, #A=3#

Coefficients of #x^2#

#4=A+B#, #=>#, #B=4-A=4-3=1#

Coefficients of #x#

#-1=C#

Therefore,

#(4x^2-x+12)/(x^3+4x)=3/(x)+(x-1)/(x^2+4)#

So,
#int((4x^2-x+12dx)/(x^3+4x)=int(3dx)/(x)+int((x-1)dx)/(x^2+4)#

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

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

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