Question

$$\underset{0}{\overset{1}{\int }}\frac{1-x^2}{x^4+x^2+1}dx$$

Answer 1

$$LetI=\int \frac{1-x^2}{x^4+x^2+1}dx$$
$$=-\int \frac{x^2-1}{x^4+x^2+1}dx$$
$$=-\int \frac{1-\frac{1}{x^2}}{x^2+1+\frac{1}{x^2}}dx$$
$$=-\int \frac{1-\frac{1}{x^2}}{x^2+2+\frac{1}{x^2}-1}dx$$
$$=-\int \frac{1-\frac{1}{x^2}}{{(x+\frac{1}{x})}^2-1}dx$$
$$Let,x+\frac{1}{x}=t$$
$$\Rightarrow (1-\frac{1}{x^2})dx=dt$$
$$\text{Then integral becomes},$$
$$I=-\int \frac{1}{t^2-1}dt$$
$$=-\frac{1}{2}\log \left|\frac{t-1}{t+1}\right|$$
$$=\frac{1}{2}\log \left|\frac{t+1}{t-1}\right|$$
$$=\frac{1}{2}\log \left|\frac{x+\frac{1}{x}+1}{x+\frac{1}{x}-1}\right|$$
$$=\frac{1}{2}\log \left|\frac{x^2+x+1}{x^2-x+1}\right|$$
$$i.e.,\int \frac{1-x^2}{x^4+x^2+1}dx=\frac{1}{2}\log \left|\frac{x^2+x+1}{x^2-x+1}\right|$$
$$\Rightarrow {\int }_0^1\frac{1-x^2}{x^4+x^2+1}dx={[\frac{1}{2}\log \left|\frac{x^2+x+1}{x^2-x+1}\right|]}_0^1$$
$$=\frac{1}{2}\log 3$$