Question Number 66066 by Kunal12588 last updated on 08/Aug/19
![Integrate ∫(dx/(ax^2 +bx+c)) y = ax^2 +bx+c y′=(dy/dx)=2ax+b d = (√(b^2 −4ac)) d′ = (√(−d)) Case 1. d^2 < 0 I=(2/d′)tan^(−1) ((y′)/d′)+C [tan^(−1) α=arctan α] Case 2. d^2 =0 I=((−2)/(y′))+C Case 3. d^2 > 0 I=(1/d)ln∣((y′−d)/(y′+d))∣+C [ln a = log_e a]](https://www.tinkutara.com/question/Q66066.png)
$${Integrate}\:\int\frac{{dx}}{{ax}^{\mathrm{2}} +{bx}+{c}} \\ $$$${y}\:=\:{ax}^{\mathrm{2}} +{bx}+{c} \\ $$$${y}'=\frac{{dy}}{{dx}}=\mathrm{2}{ax}+{b} \\ $$$${d}\:=\:\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}} \\ $$$${d}'\:=\:\sqrt{−{d}} \\ $$$${Case}\:\mathrm{1}.\:{d}^{\mathrm{2}} \:<\:\mathrm{0} \\ $$$${I}=\frac{\mathrm{2}}{{d}'}{tan}^{−\mathrm{1}} \frac{{y}'}{{d}'}+{C}\:\:\:\left[{tan}^{−\mathrm{1}} \alpha={arctan}\:\alpha\right] \\ $$$${Case}\:\mathrm{2}.\:{d}^{\mathrm{2}} =\mathrm{0} \\ $$$${I}=\frac{−\mathrm{2}}{{y}'}+{C} \\ $$$${Case}\:\mathrm{3}.\:{d}^{\mathrm{2}} \:>\:\mathrm{0} \\ $$$${I}=\frac{\mathrm{1}}{{d}}\mathrm{ln}\mid\frac{{y}'−{d}}{{y}'+{d}}\mid+{C}\:\:\:\left[\mathrm{ln}\:{a}\:=\:\mathrm{log}_{{e}} \:{a}\right] \\ $$