If (dx)/((x^2+x+1)^2)=a^-1(2x+1√3)+b2x+1x^2+x+1+C, x0, where C is the constant of integra…

Question

If $\displaystyle\int\frac{dx}{(x^2+x+1)^2}=a	an^{-1}\left(\dfrac{2x+1}{\sqrt3}ight)+b\cdot\dfrac{2x+1}{x^2+x+1}+C$, $x>0$, where C is the constant of integration, then the value of $9(\sqrt3a+b)$ is equal to

Answer: 15

Accepted Answer

$$\int\frac{dx}{\left(\left(x+\frac12\right)^2+\left(\frac{\sqrt3}2\right)^2\right)^2}$$

Let $x+\dfrac12=\dfrac{\sqrt3}{2}\tan\theta\Rightarrow dx=\dfrac{\sqrt3}{2}\sec^2\theta\,d\theta$:
$$\int\frac{\frac{\sqrt3}2\sec^2\theta\,d\theta}{\frac9{16}\sec^4\theta}=\frac{8}{3\sqrt3}\int\cos^2\theta\,d\theta=\frac{4}{3\sqrt3}\left(\theta+\frac{\sin2\theta}{2}\right)+C$$

Converting back to $x$: $\theta=\tan^{-1}\dfrac{2x+1}{\sqrt3}$ and $\sin2\theta=\dfrac{2\tan\theta}{1+\tan^2\theta}$, giving
$$I=\frac{4}{3\sqrt3\cdot 2}\tan^{-1}\frac{2x+1}{\sqrt3}\cdot 2 + \frac13\cdot\frac{2x+1}{x^2+x+1}+C$$

Comparing: $a=\dfrac{4}{3\sqrt3},\ b=\dfrac13$. So $9(\sqrt3a+b)=9\left(\dfrac43+\dfrac13\right)=15$

Solutions · 1