Question

$\int\limits^1_0 dx/x²+x+1 ,Evaluate the given integral expression:$

Answer 1

HELLO DEAR,

GIVEN:-
$\sf{\int\limits^1_0 dx/(x^2 + x + 1)}$

$\sf{\implies \int\limits^1_0 dx/[x^2 + x + (1/4) - (1/4) + 1]}$

$\sf{\implies \int\limits^1_0 dx/[(x + 1/2)^2 + 3/4]}$

$\sf{\implies [\frac{1}{\frac{\sqrt{3}}{2}}tan^{-1}(\frac{2x + 1}{\sqrt{3}})]^1_0}$

$\sf{\implies \frac{1}{\frac{\sqrt{3}}{2}}*[tan^{-1}(\sqrt{3}) - tan^{-1}(1/\sqrt{3})]}$

$\sf{\implies \frac{2}{\sqrt{3}}[\pi/3 - \pi/6]}$

$\sf{\implies \frac{\pi}{3\sqrt{3}}}$

I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

we have to find the value of $\int\limits^1_0{\frac{1}{x^2+x+1}}\,dx$

first of all, we have to resolve x² + x + 1.
x² + x + 1 = x² + x + 1/4 + 3/4
= x² +2.1/2.x + (1/2)² + (√3/2)²
= (x + 1/2)² + (√3/2)²

now, 1/(x² + x + 1) = 1/{(x + 1/2)² + (√3/2)²}

so, $\int\limits^1_0{\frac{1}{(x+1/2)^2+(\sqrt{3}/2)^2}}\,dx$

now, use the formula,
$\int{\frac{1}{x^2+a^2}}\,dx=\frac{1}{a}tan^{-1}\frac{x}{a}$

so, $=\left[\begin{array}{c}\frac{2}{\sqrt{3}}tan^{-1}\frac{x+1/2}{\sqrt{3}/2}\end{array}\right]^1_0$

$=\frac{2}{\sqrt{3}}[tan^{-1}\sqrt{3}-tan^{-1}\frac{1}{\sqrt{3}}]$

$=\frac{2}{\sqrt{3}}[\frac{\pi}{3}-\frac{\pi}{6}]$

$=\frac{\pi}{3\sqrt{3}}$