Question

If ${\int }_{-5}^{-2}{\left(\frac{x^2-x}{x^3-3x+1}\right)}^{2}dx+{\int }_{1/6}^{1/3}{\left(\frac{x^2-x}{x^3-3x+1}\right)}^{2}dx+{\int }_{6/5}^{3/2}{\left(\frac{x^2-x}{x^3-3x+1}\right)}^{2}dx=\frac{p}{q}$, where $p$ and $q$ are co-primes, then $\frac{6q-p}{6}$ is equal to

Answer 1

Part-I

${\int }_{-5}^{-2}{\left(\frac{x^2-x}{x^3-3x+1}\right)}^{2}dx$

substitute $x=-u$

$={\int }_5^2{\left(\frac{u^2+u}{-u^3+3u+1}\right)}^2(-du)$

$={\int }_2^5{\left(\frac{u^2+u}{u^3-3u-1}\right)}^{2}du$

Part-II

${\int }_{\frac{1}{6}}^{\frac{1}{3}}{\left(\frac{x^2-x}{x^3-3x+1}\right)}^{2}dx$

Put $t=\frac{1}{x}$

$={\int }_6^3-\frac{1}{t^2}{\left(\frac{\frac{1}{t^2}-\frac{1}{t}}{\frac{1}{t^3}-3\frac{1}{t}+1}\right)}^{2}dt$

$={\int }_3^6{\left(\frac{t-1}{t^3-3t^2+1}\right)}^{2}dt$

Put $u=t-1$

$={\int }_2^5{\left(\frac{u}{u^3-3u-1}\right)}^{2}du$

Part-III

${\int }_{\frac{6}{5}}^{\frac{3}{2}}{\left(\frac{x^2-x}{x^3-3x+1}\right)}^{2}dx$

Put $x=\frac{u+1}{u}$

$={\int }_5^2-\frac{1}{u^2}{\left(\frac{\frac{u+1}{u}\times \frac{1}{u}}{{\left(\frac{u+1}{u}\right)}^3-3\left(\frac{u+1}{u}\right)+1}\right)}^{2}du$

$={\int }_2^5{\left(\frac{u+1}{u^3-3u-1}\right)}^{2}du$

Combining all 3 parts, we have

${\int }_2^5{\left(\frac{1}{u^3-3u-1}\right)}^2\left(u^2\right(u+1{)}^2+u^2+(u+1{)}^2)du$

$={\int }_2^5\left(\frac{u-2}{u^3-3u-1}\right)-\left(\frac{5u+1}{(u^3-3u-1{)}^2}\right)du=\frac{23}{4}$

$\therefore \frac{6q-p}{6}=\frac{1}{6}$