Question

integration of 1 by 1 + x square + x​

Answer 1

Given to find,

$\displaystyle\longrightarrow\sf{\int\dfrac{1}{x^2+x+1}\ dx}$

By factorising the denominator $\sf{x^2+x+1}$ we get complex roots so let,

$\displaystyle\longrightarrow\sf{\dfrac{1}{x^2+x+1}\ dx=\dfrac{k}{u^2+1}\ du}$

where $\sf{k}$ is a constant and $\sf{u}$ is a function in $\sf{x,}$ because we have to bring the integral as,

$\displaystyle\longrightarrow\sf{\int\dfrac{1}{x^2+x+1}\ dx=\int\dfrac{k}{u^2+1}\ du\quad\quad\dots(1)}$

$\displaystyle\longrightarrow\sf{\int\dfrac{1}{x^2+x+1}\ dx=k\tan^{-1}(u)+c\quad\quad\dots(2)}$

So,

$\displaystyle\longrightarrow\sf{\int\dfrac{1}{x^2+x+1}\ dx=\int\dfrac{1}{x^2+x+\dfrac{1}{4}+\dfrac{3}{4}}\ dx}$

$\displaystyle\longrightarrow\sf{\int\dfrac{1}{x^2+x+1}\ dx=\int\dfrac{1}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\ dx}$

Multiplying $\sf{\dfrac{4}{3}}$ on both the numerator and the denominator,

$\displaystyle\longrightarrow\sf{\int\dfrac{1}{x^2+x+1}\ dx=\int\dfrac{\dfrac{4}{3}}{\left[\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]\dfrac{4}{3}}\ dx}$

$\displaystyle\longrightarrow\sf{\int\dfrac{1}{x^2+x+1}\ dx=\dfrac{4}{3}\int\dfrac{1}{\left[\left(x+\dfrac{1}{2}\right)\dfrac{2}{\sqrt3}\right]^2+1}\ dx}$

$\displaystyle\longrightarrow\sf{\int\dfrac{1}{x^2+x+1}\ dx=\dfrac{4}{3}\int\dfrac{1}{\left(\dfrac{2x}{\sqrt3}+\dfrac{1}{\sqrt3}\right)^2+1}\ dx}$

$\displaystyle\longrightarrow\sf{\int\dfrac{1}{x^2+x+1}\ dx=\dfrac{4}{3}\int\dfrac{1}{\left(\dfrac{2x+1}{\sqrt3}\right)^2+1}\ dx\quad\quad\dots(3)}$

Comparing the denominator with $\sf{u^2+1}$ then we get,

$\displaystyle\longrightarrow\sf{u=\dfrac{2x+1}{\sqrt3}}$

$\displaystyle\longrightarrow\sf{\dfrac{du}{dx}=\dfrac{2}{\sqrt3}}$

$\displaystyle\longrightarrow\sf{dx=\dfrac{\sqrt3}{2}\ du}$

Then (3) becomes,

$\displaystyle\longrightarrow\sf{\int\dfrac{1}{x^2+x+1}\ dx=\dfrac{4}{3}\int\dfrac{1}{u^2+1}\cdot\dfrac{\sqrt3}{2}\ du}$

$\displaystyle\longrightarrow\sf{\int\dfrac{1}{x^2+x+1}\ dx=\dfrac{2}{\sqrt3}\int\dfrac{1}{u^2+1}\ du\quad\quad\dots(4)}$

From (1) and (4) we get,

$\displaystyle\longrightarrow\sf{k=\dfrac{2}{\sqrt3}}$

Hence (2) becomes,

$\displaystyle\longrightarrow\sf{\underline{\underline{\int\dfrac{1}{x^2+x+1}\ dx=\dfrac{2}{\sqrt3}\tan^{-1}\left(\dfrac{2x+1}{\sqrt3}\right)+c}}}$