Question

$\int \frac{{\left(x-1\right)}^2}{x^4+x^2+1}dx$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \frac{{\left(x-1\right)}^{2}dx}{x^4+x^2+1} \\ =\int \left(\frac{x^2-2x+1}{x^4+x^2+1}\right)dx \\ =\int \left(\frac{x^2+1}{x^4+x^2+1}\right)dx-\int \frac{2xdx}{x^4+x^2+1} \\ ={\mathrm{I}}_1-{\mathrm{I}}_2 \\ \mathrm{where}, \\ {\mathrm{I}}_1=\int \frac{\left(x^2+1\right)dx}{x^4+x^2+1} \\ {\mathrm{I}}_2=\int \frac{2xdx}{x^4+x^2+1} \\ \mathrm{Now}, \\ {\mathrm{I}}_1=\int \left(\frac{x^2+1}{x^4+x^2+1}\right)dx \\ \mathrm{Dividing}\mathrm{numerator}\mathrm{and}\mathrm{denominator}\mathrm{by}{x}^2 \\ {\mathrm{I}}_1=\int \left(\frac{1+{\displaystyle \frac{1}{x^2}}}{x^2+{\displaystyle \frac{1}{x^2}}+1}\right)dx \\ {\mathrm{I}}_1=\int \frac{\left(1+{\displaystyle \frac{1}{x^2}}\right)dx}{x^2+{\displaystyle \frac{1}{x^2}}-2+3} \\ {\mathrm{I}}_1=\int \frac{\left(1+{\displaystyle \frac{1}{x^2}}\right)dx}{{\left(x-{\displaystyle \frac{1}{x}}\right)}^2+{\left(\sqrt{3}\right)}^2} \\ \mathrm{Putting}x-\frac{1}{x}=t \\ \Rightarrow \left(1+\frac{1}{x^2}\right)dx=dt \\ \therefore {\mathrm{I}}_1=\int \frac{dt}{t^2+{\left(\sqrt{3}\right)}^2} \\ =\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{t}{\sqrt{3}}\right)+C_1 \\ =\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{x-{\displaystyle \frac{1}{x}}}{\sqrt{3}}\right)+C_1 \\ =\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{x^2-1}{\sqrt{3}x}\right)+C_1 \\ \mathrm{And} \\ {\mathrm{I}}_2=\int \frac{2xdx}{x^4+x^2+1} \\ \mathrm{Putting}{x}^2=t \\ \Rightarrow 2xdx=dt \\ {\mathrm{I}}_2=\int \frac{dt}{t^2+t+1} \\ =\int \frac{dt}{t^2+t+{\left({\displaystyle \frac{1}{2}}\right)}^2-{\left({\displaystyle \frac{1}{2}}\right)}^2+1} \\ =\int \frac{dt}{{\left(t+{\displaystyle \frac{1}{2}}\right)}^2+{\left({\displaystyle \frac{\sqrt{3}}{2}}\right)}^2} \\ =\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{t+{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{\sqrt{3}}{2}}}\right)+{\mathrm{C}}_2 \\ =\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2t+1}{3}\right)+{\mathrm{C}}_2 \\ \therefore I=\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{x^2-1}{\sqrt{3}x}\right)-\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x^2+1}{\sqrt{3}}\right)+\mathrm{C} \end{gathered}$$