How do you integrate #int (x^3 +x^2+2x+1) / [(x^2+1) (x^2+2)]# using partial fractions?

1 Answer
Feb 22, 2016

#1/2 ln (x^2 +1) +1/sqrt(2) arctan(x/sqrt2) +C#

#1/2 ln(x^2+1) +sqrt2 arctan(x/sqrt2)+C#

Explanation:

Note: If the denominator of the partial fraction is not factorable, always remember the degree of the numerator is one less. Like the problem

Step 1: Find the equivalent partial fraction

#(x^3+x^2+2x+1)/((x^2+1)(x^2+2)) = (Ax+B)/(x^2 +1) + (Cx+D)/(x^2+2)#

Step 2 Solve the partial fraction by multiply by Least common denominator to get

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

Step 3: Multiply/expand/foil the right hand side to get

#x^3 + x^2 + 2x+1= Ax^3 + 2Ax + Bx^2 + 2B + Cx^3 + Cx + Dx^2 + D#

Step 4: Set up the system of equation, using the corresponding coefficient for the corresponding terms

#x^3 " "term" " " 1= A + C#
#x^2 " "term" " " 1= B+ D#
#x " "term" " " 2= 2A +C#
#x^0 " "term" " " 1= 2B+ D#

Step 5: Solve the system of equation
#-2(1 = A + C) => -2 = -2A + 2C#

#+(2 = 2A + C)#
#0 = 3C => C= 0# , #A= 1#

#-1(2B +D = 1) => -2B -D= -1#

#+ (B +D = 1)#
#-B= 0=> B= 0, D= 1#

Step 6: Write the equivalent partial fraction for the integral like this

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

Step 7 Integrate the integral

the first integral can be done by u substitution like this

let #u= x^2 +1#
#du = 2xdx => (du)/2 = xdx#
Note: #int(dx)/(x) = ln|x| +C " " "or" " " log|x|+C#

The second integral is #arctan# , remember #int 1/(a^2+u^2)du =1/a arctant u/a +C #

Answer:

#1/2 ln (x^2 +1) +1/sqrt(2) arctan(x/sqrt2) +"Constant"#

I let my constant value to be #C#

this is another answer: #1/2 ln (x^2 +1) +1/sqrt(2) arctan(x/sqrt2) +C# or

#1/2 ln(x^2+1) +sqrt2 arctan(x/sqrt2)+C#