How do you find #int (x+4)/(x^2+2x+5) dx# using partial fractions?
1 Answer
Explanation:
Since we are requested to use partial fractions to solve this, I guess the hope is that we can simplify the integrand.
However, the quadratic
They say fools rush in where angels fear to tread, so here I go...
#x^2+2x+5 = (x+1)^2+2^2#
#color(white)(x^2+2x+5) = (x+1)^2-(2i)^2#
#color(white)(x^2+2x+5) = ((x+1)-2i)((x+1)+2i)#
#color(white)(x^2+2x+5) = (x+1-2i)(x+1+2i)#
So:
#(x+4)/(x^2+2x+5) = A/(x+1-2i)+B/(x+1+2i)#
Using Heaviside's cover up method, we find:
#A = (color(blue)(-1+2i)+4)/(color(blue)(-1+2i)+1+2i) = (3+2i)/(4i)= 1/2-3/4i#
#B = bar(A) = 1/2+3/4i#
So:
#int (x+4)/(x^2+2x+5) dx#
#= int (1/2-3/4i) * 1/(x+1-2i) + (1/2+3/4i) * 1/(x+1+2i) dx#
#= (1/2-3/4i) ln (x+1-2i) + (1/2+3/4i) ln (x+1+2i) + C#
#= 1/2(ln (x+1-2i) + ln (x+1+2i))+3/4 i (ln(x+1+2i)-ln(x+1-2i)) + C#
#= 1/2 ln (x^2+2x+5)+3/4 i (ln(x+1+2i)-ln(x+1-2i)) + C#
Use:
#tan^(-1)(z) = 1/2i(ln(1-iz)-ln(1+iz))#
So:
#3/4 i (ln(x+1+2i)-ln(x+1-2i))#
#= 3/4 i (ln(1+i(2/(x+1))) - ln(1-i(2/(x+1))))#
#= -3/2 1/2 i (ln(1-i(2/(x+1))) - ln(1+i(2/(x+1))))#
#= -3/2 tan^(-1)(2/(x+1))#
#= 3/2 tan^(-1)((x+1)/2) + (3pi)/4#
Hence:
#int (x+4)/(x^2+2x+5) dx = 1/2 ln (x^2+2x+5) + 3/2 tan^(-1)((x+1)/2) + C#