Question

$\int \frac{x^2-3x+1}{x^4+x^2+1}dx$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \left(\frac{x^2-3x+1}{x^4+x^2+1}\right)dx \\ =\int \frac{\left(x^2+1\right)dx}{x^4+x^2+1}-3\int \frac{xdx}{x^4+x^2+1}.....\left(1\right) \\ ={\mathrm{I}}_1-3{\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{xdx}{x^4+x^2+1} \\ {I}_1=\int \left(\frac{x^2+1}{x^4+x^2+1}\right)dx \\ \mathrm{Dividing}\mathrm{numerator}\&\mathrm{denominator}\mathrm{by}{x}^2 \\ {\mathrm{I}}_1=\int \frac{\left(1+{\displaystyle \frac{1}{x^2}}\right)dx}{x^2+{\displaystyle \frac{1}{x^2}}+1} \\ =\int \frac{\left(1+{\displaystyle \frac{1}{x^2}}\right)dx}{x^2+{\displaystyle \frac{1}{x^2}}-2+3} \\ =\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{Let}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} \\ {\mathrm{I}}_1=\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{t}{\sqrt{3}}\right)+{\mathrm{C}}_1 \\ {\mathrm{I}}_1=\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{x-{\displaystyle \frac{1}{x}}}{\sqrt{3}}\right)+{\mathrm{C}}_1.....\left(2\right) \\ {\mathrm{I}}_2=\int \frac{xdx}{x^4+x^2+1} \\ \mathrm{Putting}{x}^2=t \\ \Rightarrow 2xdx=dt \\ \Rightarrow xdx=\frac{dt}{2} \\ \therefore {\mathrm{I}}_2=\frac{1}{2}\int \frac{dt}{t^2+t+1} \\ =\frac{1}{2}\int \frac{dt}{t^2+t+{\displaystyle \frac{1}{4}}+{\displaystyle \frac{3}{4}}} \\ =\frac{1}{2}\int \frac{dt}{{\left(t+{\displaystyle \frac{1}{2}}\right)}^2+{\left({\displaystyle \frac{\sqrt{3}}{2}}\right)}^2} \\ =\frac{1}{{\displaystyle \frac{\sqrt{3}}{2}}}\times \frac{1}{2}\left[{\tan }^{-1}\left(\frac{t+{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{\sqrt{3}}{2}}}\right)\right]+{\mathrm{C}}_2 \\ =\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{2t+1}{\sqrt{3}}\right)+{\mathrm{C}}_2 \\ {\mathrm{I}}_2=\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x^2+1}{\sqrt{3}}\right)+{\mathrm{C}}_2...\left(3\right) \\ \mathrm{From}\mathrm{equating}\left(1\right),\left(2\right)\mathrm{and}\left(3\right)\mathrm{we}\mathrm{have} \\ I=\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{x-{\displaystyle \frac{1}{x}}}{\sqrt{3}}\right)+{\mathrm{C}}_1-3\times \left[\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{2{\mathrm{x}}^2+1}{\sqrt{3}}\right)+{\mathrm{C}}_2\right] \\ =\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{x^2-1}{\sqrt{3}x}\right)-\sqrt{3}{\tan }^{-1}\left(\frac{2x^2+1}{\sqrt{3}}\right)+\mathrm{C}\mathrm{where}\mathrm{C}={\mathrm{C}}_1+3{\mathrm{C}}_2 \end{gathered}$$