Find #int \ (x^2+x+1)/(x^2(x+2)) \ dx # using partial fractions?
1 Answer
# int \ (x^2+x+1)/(x^2(x+2)) \ dx = 1/4ln|x| - 1/(2x)+ 3/4 ln |x+2|+C #
Explanation:
We seek:
# I = int \ (x^2+x+1)/(x^2(x+2)) \ dx #
We can decompose the integrand into partial fractions:
# (x^2+x+1)/(x^2(x+2)) -= A/x + B/x^2 + C/(x+2) #
# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = ( Ax(x+2) + B(x+2) + Cx^2 ) / (x^2(x+2)) #
Leading to:
# x^2+x+1 -= Ax(x+2) + B(x+2) + Cx^2 #
Where
Put
# x = 0 => 1= 2B => B =1/2#
Put# x = -2 => 3 = 4C => C=3/4 #
Coeff#(x^2): 1 = A+C => A = 1/4#
Hence, we can now write the integral as:
# I = int \ (1/4)/x + (1/2)/x^2 + (3/4)/(x+2) \ dx #
# \ \ = 1/4 int \ 1/x \ dx + 1/2int \ 1/x^2 \ dx + 3/4 \ int 1/(x+2) \ dx #
# \ \ = 1/4ln|x| +1/2 x^(-1)/(-1) + 3/4 ln |x+2|+C #
# \ \ = 1/4ln|x| - 1/(2x)+ 3/4 ln |x+2|+C #
