∫0∞(4x2+17x+4)(x2+4x+5)x2−32x−69dx
However, I'll still wait in the car....∫0∞(4x2+17x+4)(x2+4x+5)x2−32x−69dx
=∫0∞(4x+1)(x+4)(x+4)(x+1)x2−32x−69dx
=∫0∞(4x+1)(x+4)2(x+1)x2−32x−69dx
Dr. Flim-Flam said:Why not 2x/(x^2+4x+5) + 4/(x^2+4x+5) -4/(4x+1) - 1/(x+4).
Then the antiderivative is ln|x^2+4x+5| + 4arctan(x+2)- 4arctan(x+2) - ln(4x+1) -ln(x+4|, which
reduces to -ln(5) over the interval 0 to infinity.
Note: In the comment immediately above my comment the denominator doesn't factor.
x^2+4x+5 is prime.<--- Correct
I stand corrected - that was a mistake