Mark M. answered 03/17/24
Retired math prof. Calc 1, 2 and AP Calculus tutoring experience.
∫ [ (3x) / (x2+8x+21) ]dx + ∫ [2 / (x2+8x+21) ]dx
= (3/2)∫ [(2x + 8 - 8) / (x2+8x+21)]dx + ∫ [2 / ((x+4)2 + 5)]dx
= (3/2)∫ [(2x+8) / (x2+8x+21)]dx - 10∫ [dx / ((x+4)2 + 5) ]
= (3/2) ln l x2+8x+21 l - 2∫ [dx / ({(x+4)/√5)}2 + 1)] Let u = (x+4)/√5. Then du = (1/√5)dx.
= (3/2) ln l x2+8x+21 l - 2√5 Tan-1((x+4)/√5) + C