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