Question

${\int }_{\frac{1}{2}}^2\frac{{\tan }^{-1}x}{x^2-x+1}dx=\frac{{\pi }^2\sqrt{3}}{2a}$. Find a

Answer 1

Take $x=\frac{1}{y}$

$\to$ $dx=d\left[\frac{1}{y}\right]\Rightarrow dx=\frac{-1}{y^2}dy.$

$Therefore,$

${\int }_{1/2}^2\frac{{\tan }^{-1}x}{x^2-x+1}dx={\int }_2^{1/2}\frac{{\tan }^{-1}\frac{1}{y}}{\frac{y^2-y+1}{y^2}}\times \frac{-1}{y^2}dy$

$\to {\int }_{1/2}^2\frac{{\tan }^{-1}x}{x^2-x+1}dx=-{\int }_2^{1/2}\frac{{\cot }^{-1}y}{y^2-y+1}dy$

$\to {\int }_{1/2}^2\frac{{\tan }^{-1}x}{x^2-x+1}dx={\int }_{1/2}^2\frac{{\cot }^{-1}y}{y^2-y+1}dy$ $................1$

Using equation 1, we can write:

$2{\int }_{1/2}^2\frac{{\tan }^{-1}x}{x^2-x+1}dx={\int }_{1/2}^2\frac{{\cot }^{-1}y}{y^2-y+1}dy+{\int }_{1/2}^2\frac{{\tan }^{-1}x}{x^2-x+1}dx$

$\to 2{\int }_{1/2}^2\frac{{\tan }^{-1}x}{x^2-x+1}dx={\int }_{1/2}^2\frac{\frac{\pi }{2}}{x^2-x+1}dx$

Converting the denominator, we can write the above equation as:

$2{\int }_{1/2}^2\frac{{\tan }^{-1}x}{x^2-x+1}dx={\int }_{1/2}^2\frac{\frac{\pi }{2}}{{[x-\frac{1}{2}]}^2+\frac{3}{4}}dx$

Because, $x^2-x+1=x^2-2.\frac{x}{2}+\frac{1}{4}+\frac{3}{4}$

We can also write it as:

$2{\int }_{1/2}^2\frac{{\tan }^{-1}x}{x^2-x+1}dx={\int }_{1/2}^2\frac{\frac{\pi }{2}}{{[x-\frac{1}{2}]}^2+{\left[\sqrt{\frac{3}{4}}\right]}^2}dx$

Now using the below formula:

$\int \frac{1}{x^2+a^2}dx=\frac{1}{a}{\tan }^{-1}\frac{x}{a}+c$ we will get:

$a=9$