Question

Integrate
1/(x+1)sqrt(x^2+x+1)​

Answer 1

Given to evaluate,

$\small\text{ \displaystyle\longrightarrow I=\int\dfrac{dx}{(x+1)\sqrt{x^2+x+1}} }$

Completing the square in x² + x + 1,

$\small\text{ \displaystyle\longrightarrow I=\int\dfrac{dx}{(x+1)\sqrt{x^2+x+\dfrac{1}{4}+\dfrac{3}{4}}} }$

$\small\text{ \displaystyle\longrightarrow I=\int\dfrac{dx}{(x+1)\sqrt{\left(x+\dfrac{1}{2}\right)^2+\left(\dfrac{\sqrt3}{2}\right)^2}}\quad\dots(1) }$

Substitute,

$\small\longrightarrow x+\dfrac{1}{2}=\dfrac{\sqrt3}{2}\,\tan\theta$

Then we get the following.

$\small\longrightarrow\theta=\tan^{-1}\left(\dfrac{2x+1}{\sqrt3}\right)$

$\small\text{ \longrightarrow\sqrt{\left(x+\dfrac{1}{2}\right)^2+\left(\dfrac{\sqrt3}{2}\right)^2}=\dfrac{\sqrt3}{2}\,\sec\theta }$

$\small\longrightarrow x+1=\dfrac{\sqrt3}{2}\,\tan\theta+\dfrac{1}{2}$

$\small\longrightarrow dx=\dfrac{\sqrt3}{2}\,\sec^2\theta\,d\theta$

Then (1) becomes,

$\small\text{ \displaystyle\longrightarrow I=\int\dfrac{\dfrac{\sqrt3}{2}\sec^2\theta\,d\theta}{\left(\dfrac{\sqrt3}{2}\tan\theta+\dfrac{1}{2}\right)\dfrac{\sqrt3}{2}\sec\theta} }$

$\small\text{ \displaystyle\longrightarrow I=\int\dfrac{\sec\theta\,d\theta}{\dfrac{\sqrt3}{2}\tan\theta+\dfrac{1}{2}} }$

$\small\text{ \displaystyle\longrightarrow I=\int\dfrac{d\theta}{\cos\theta\left(\dfrac{\sqrt3}{2}\tan\theta+\dfrac{1}{2}\right)} }$

$\small\text{ \displaystyle\longrightarrow I=\int\dfrac{d\theta}{\dfrac{\sqrt3}{2}\sin\theta+\dfrac{1}{2}\cos\theta} }$

$\small\text{ \displaystyle\longrightarrow I=\int\dfrac{d\theta}{\sin\theta\cos\left(\dfrac{\pi}{6}\right)+\cos\theta\sin\left(\dfrac{\pi}{6}\right)} }$

$\small\text{ \displaystyle\longrightarrow I=\int\dfrac{d\theta}{\sin\left(\theta+\dfrac{\pi}{6}\right)} }$

$\small\displaystyle\longrightarrow I=\int\csc\left(\theta+\dfrac{\pi}{6}\right)\,d\theta$

$\small\displaystyle\longrightarrow I=\log\left|\csc\left(\theta+\dfrac{\pi}{6}\right)-\cot\left(\theta+\dfrac{\pi}{6}\right)\right|+C$

$\small\text{ \displaystyle\longrightarrow I=\log\left|\dfrac{1-\cos\left(\theta+\dfrac{\pi}{6}\right)}{\sin\left(\theta+\dfrac{\pi}{6}\right)}\right|+C }$

Since 1 - cos(2α) = 2sin²(α) and sin(2α) = 2sin(α)cos(α),

$\small\text{ \displaystyle\longrightarrow I=\log\left|\dfrac{2\sin^2\left(\dfrac{\theta}{2}+\dfrac{\pi}{12}\right)}{2\sin\left(\dfrac{\theta}{2}+\dfrac{\pi}{12}\right)\cos\left(\dfrac{\theta}{2}+\dfrac{\pi}{12}\right)}\right|+C }$

$\small\displaystyle\longrightarrow I=\log\left|\tan\left(\dfrac{\theta}{2}+\dfrac{\pi}{12}\right)\right|+C$

Undoing substitution for θ,

$\small\text{ \displaystyle\longrightarrow\underline{\underline{I=\log\left|\tan\left(\dfrac{1}{2}\tan^{-1}\left(\dfrac{2x+1}{\sqrt3}\right)+\dfrac{\pi}{12}\right)\right|+C}} }$