How do you find #int (x^3+x^2+2x+1)/((x^2+1)(x^2+2)) dx# using partial fractions?

Question

Ezekiel Alexander · Accepted Answer

#1/2ln|x^2+1|+1/sqrt2tan^-1(x/sqrt2)+c#

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

Rearranging the terms in numerator,

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

Simplifying according to required factors in denominator,

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

#=int(xcancel((x^2+2)))/((x^2+1)cancel((x^2+2)))dx+intcancel((x^2+1))/(cancel((x^2+1))(x^2+2))dx#

#=intx/(x^2+1)dx+int1/(x^2+2)dx#

#=1/2int(2x)/(x^2+1)dx+int1/(x^2+(sqrt2)^2)dx#

#=1/2int(d/(dx)(x^2+1))/(x^2+1)dx+int1/(x^2+(sqrt2)^2)dx#

#=1/2ln|x^2+1|+1/sqrt2tan^-1(x/sqrt2)+c#

Luke Alvarez · Answer

undefined To find the integral of (x^3+x^2+2x+1)/((x^2+1)(x^2+2)) using partial fractions, follow these steps:

1. Factor the denominator: (x^2+1)(x^2+2) = (x^2+1)(x^2+2)
2. Decompose the fraction into partial fractions:
   (x^3+x^2+2x+1)/((x^2+1)(x^2+2)) = A/(x^2+1) + B/(x^2+2)
3. Multiply both sides by the common denominator to clear fractions:
   x^3 + x^2 + 2x + 1 = A(x^2 + 2) + B(x^2 + 1)
4. Expand and collect like terms:
   x^3 + x^2 + 2x + 1 = (A + B)x^2 + 2A + B
5. Equate coefficients of corresponding powers of x:
   For x^3 term: 0 = 0
   For x^2 term: 1 = A + B
   For x term: 2 = 0
   For constant term: 1 = 2A + B
6. Solve the system of equations:
   From the second equation: A = 1 - B
   Substitute A = 1 - B into the fourth equation: 1 = 2(1 - B) + B
   Solve for B: B = 1/2
   Substitute B = 1/2 into A = 1 - B: A = 1 - 1/2 = 1/2
7. Rewrite the integral with the partial fractions:
   ∫((1/2)/(x^2+1) + (1/2)/(x^2+2)) dx
8. Integrate each term:
   ∫(1/2)/(x^2+1) dx = (1/2) arctan(x) + C1
   ∫(1/2)/(x^2+2) dx = (1/2√2) arctan(x/√2) + C2
9. Combine the integrals and constants:
   ∫(x^3+x^2+2x+1)/((x^2+1)(x^2+2)) dx = (1/2) arctan(x) + (1/2√2) arctan(x/√2) + C
   where C is the constant of integration.