Question

Integrate $\int \frac{x}{x^4+x^2+1}dx$.

Answer 1

Solve the given integral

Given, $\int \frac{x}{x^4+x^2+1}dx$

Put , $x^2=t$

Then $2xdx=dt$

Now,

$$\begin{gathered} \int \frac{x}{{\left(x^2\right)}^2+x^2+1}dx \\ =\frac{1}{2}\int \frac{1}{{\left(t\right)}^2+t+1}dt \\ =\frac{1}{2}\int \frac{1}{t^2+t+1+{\left({\displaystyle \frac{1}{2}}\right)}^2-{\left({\displaystyle \frac{1}{2}}\right)}^2}dt \\ =\frac{1}{2}\int \frac{1}{t^2+t+1+{\left({\displaystyle \frac{1}{2}}\right)}^2-{\displaystyle \frac{1}{4}}}dt \\ =\frac{1}{2}\int \frac{1}{t^2+t+{\left({\displaystyle \frac{1}{2}}\right)}^2+{\displaystyle \frac{3}{4}}}dt \end{gathered}$$

$$\begin{gathered} =\frac{1}{2}\int \frac{dt}{\left[{\left(t+{\displaystyle \frac{1}{2}}\right)}^2+\left({\displaystyle \frac{3}{4}}\right)\right]} \\ =\frac{1}{2}\int \frac{dt}{\left[{\left(t+{\displaystyle \frac{1}{2}}\right)}^2+{\left({\displaystyle \frac{\sqrt{3}}{2}}\right)}^2\right]} \end{gathered}$$

$$\begin{gathered} =\frac{1}{2}\times \left(\frac{2}{\sqrt{3}}\right){\tan }^{-1}\left[\frac{\left(t+{\displaystyle \frac{1}{2}}\right)}{{\displaystyle \frac{\sqrt{3}}{2}}}\right]+C\left[\because \int \frac{dx}{x^2+a^2}=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)\right] \\ =\left(\frac{1}{\sqrt{3}}\right){\tan }^{-1}\left(\frac{\left(2t+1\right)}{\sqrt{3}}\right)+C \\ =\left(\frac{1}{\sqrt{3}}\right){\tan }^{-1}\left(\frac{2x^2+1}{\sqrt{3}}\right)+C \end{gathered}$$

Hence , $\int \frac{x}{x^4+x^2+1}dx$ $=\left(\frac{1}{\sqrt{3}}\right){\tan }^{-1}\left(\frac{2x^2+1}{\sqrt{3}}\right)+C$.