If x1+ x + ^2x,dx = x - KA^-1K x + 1A + C, (C is a constant of integration), then the orde…

Question

If $\displaystyle\int \dfrac{	an x}{1+	an x + 	an^2x}\,dx = x - \dfrac{K}{\sqrt{A}}	an^{-1}\dfrac{K	an x + 1}{\sqrt{A}} + C$, (C is a constant of integration), then the ordered pair $(K,A)$ is equal to

Accepted Answer

Let $I = \int \dfrac{	an x}{1+	an x+	an^2x}\,dx$

$I = \int \dfrac{	an x + 1 + 	an^2x}{1+	an x+	an^2x}\,dx - \int \dfrac{1+	an^2x}{1+	an x+	an^2x}\,dx = x - \int \dfrac{\sec^2x\,dx}{1+	an x+	an^2x}$

Put $	an x = t \Rightarrow \sec^2x\,dx = dt$

$I = x - \int \dfrac{dt}{t^2+t+\frac14 + 1 - \frac14} = x - \int \dfrac{dt}{\left(t+\frac12ight)^2+\left(\frac{\sqrt3}{2}ight)^2}$

$I = x - \dfrac{2}{\sqrt3}	an^{-1}\dfrac{2	an x+1}{\sqrt3} + C$

$	herefore A = 3, K = 2$

Solutions · 1