Question

For real numbers $\alpha ,\beta ,\gamma and\delta ,$ if

$\int \frac{\left(x^2-1\right)+{\tan }^{-1}\left({\displaystyle \frac{x^2+1}{x}}\right)dx}{\left(x^4+3x^2+1\right){\tan }^{-1}\left(\frac{x^2+1}{x}\right)}=\alpha {\log }_e\left({\tan }^{-1}\left(\frac{x^2+1}{x}\right)\right)+\beta {\tan }^{-1}\left[\gamma \left(\frac{x^2+1}{x}\right)\right]+\delta {\tan }^{-1}\left(\frac{x^2+1}{x}\right)$

where $C$ is an arbitrary constant, then the value of $10(\alpha +\beta \gamma +\delta )$ is equal to

Answer 1

Step1. Divide the integration into parts:

$\begin{array}{rcl}I& =& \int \frac{\left(x^2-1\right)+{\tan }^{-1}\left({\displaystyle \frac{x^2+1}{x}}\right)dx}{\left(x^4+3x^2+1\right){\tan }^{-1}\left(\frac{x^2+1}{x}\right)}\\ & =& \int \frac{\left(x^2-1\right)dx}{\left(x^4+3x^2+1\right){\tan }^{-1}\left(\frac{x^2+1}{x}\right)}+\int \frac{{\tan }^{-1}\left(\frac{x^2+1}{x}\right)dx}{\left(x^4+3x^2+1\right){\tan }^{-1}\left(\frac{x^2+1}{x}\right)}\\ & =& \int \frac{\left(x^2-1\right)dx}{\left(x^4+3x^2+1\right){\tan }^{-1}\left(\frac{x^2+1}{x}\right)}+\int \frac{dx}{\left(x^4+3x^2+1\right)}\\ & =& I_1+I_{2\left[\mathbf{Where}\mathbf{}{\mathbf{I}}_{\mathbf{1}}\mathbf{=}\mathbf{\int }\frac{\mathbf{(}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{dx}}{\mathbf{(}{\mathbf{x}}^{\mathbf{4}}\mathbf{+}\mathbf{3}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{1}\mathbf{)}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\frac{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{1}}{\mathbf{x}}\mathbf{\right)}}\mathbf{}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{I}}_{\mathbf{2}}\mathbf{=}\mathbf{\int }\frac{\mathbf{dx}}{(x^4+3x^2+1)}\right]}\end{array}$

Step2. Calculate the value of ${\mathbf{I}}_{\mathbf{1}}$:

$\begin{array}{rcl}{\mathrm{I}}_1& =& \int \frac{({\mathrm{x}}^2-1)\mathrm{dx}}{({\mathrm{x}}^4+3{\mathrm{x}}^2+1){\tan }^{-1}\left(\frac{{\mathrm{x}}^2+1}{\mathrm{x}}\right)}\\ & =& \int \frac{({\mathrm{x}}^2-1)\mathrm{dx}}{{\mathrm{x}}^2({\mathrm{x}}^2+3+{\displaystyle \frac{1}{{\mathrm{x}}^2}}){\tan }^{-1}\left(\frac{{\mathrm{x}}^2+1}{\mathrm{x}}\right)}\\ & =& \int \frac{\left({\displaystyle \frac{{\mathrm{x}}^2-1}{{\mathrm{x}}^2}}\right)\mathrm{dx}}{({\mathrm{x}}^2+3+{\displaystyle \frac{1}{{\mathrm{x}}^2}}){\tan }^{-1}\left(\frac{{\mathrm{x}}^2+1}{\mathrm{x}}\right)}\\ & =& \int \frac{(1-{\displaystyle \frac{1}{{\mathrm{x}}^2}})\mathrm{dx}}{({\mathrm{x}}^2+2\left(\mathrm{x}\right)\left({\displaystyle \frac{1}{\mathrm{x}}}\right)+{\displaystyle \frac{1}{{\mathrm{x}}^2}+1}){\tan }^{-1}\left(\frac{{\mathrm{x}}^2+1}{\mathrm{x}}\right)}\\ & =& \int \frac{(1-{\displaystyle \frac{1}{{\mathrm{x}}^2}})\mathrm{dx}}{({\left(\mathrm{x}+{\displaystyle \frac{1}{\mathrm{x}}}\right)}^2+1^2){\tan }^{-1}(\mathrm{x}+{\displaystyle \frac{1}{\mathrm{x}}})}\\ & =& \int \frac{\mathrm{dt}}{\mathrm{t}}\left[\mathbf{Substitute}\mathbf{}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\mathbf{(}\mathbf{x}\mathbf{+}\frac{\mathbf{1}}{\mathbf{x}}\mathbf{)}\mathbf{}\mathbf{=}\mathbf{t}\mathbf{}\mathbf{\Rightarrow }\frac{\mathbf{(}\mathbf{1}\mathbf{-}{\displaystyle \frac{1}{x^2}}\mathbf{)}\mathbf{dx}}{\mathbf{(}{\left(x+{\displaystyle \frac{1}{x}}\right)}^{\mathbf{2}}\mathbf{+}{\mathbf{1}}^{\mathbf{2}}\mathbf{)}}\mathbf{=}\mathbf{dt}\right]\\ & =& {\log }_{\mathrm{e}}\left|\mathrm{t}\right|+{\mathrm{c}}_1\left[\mathbf{\because }\mathbf{\int }\frac{\mathbf{dt}}{\mathbf{t}}\mathbf{=}{\mathbf{log}}_{\mathbf{e}}\left|t\right|\right]\\ & =& {\log }_{\mathrm{e}}\left|{\tan }^{-1}(\mathrm{x}+\frac{1}{\mathrm{x}})\right|+{\mathrm{c}}_1\mathrm{\_\_\_}\left(1\right)\end{array}$

Step3.Calculate the value of ${\mathbf{I}}_{\mathbf{2}}$:

$\begin{array}{rcl}{\mathrm{I}}_2& =& \int \frac{\mathrm{dx}}{({\mathrm{x}}^4+3{\mathrm{x}}^2+1)}\\ & =& \int \frac{\mathrm{dx}}{{\mathrm{x}}^2({\mathrm{x}}^2+3+{\displaystyle \frac{1}{{\mathrm{x}}^2}})}\\ & =& \int \frac{{\displaystyle \frac{1}{{\mathrm{x}}^2}}\times {\displaystyle \frac{2}{2}}\mathrm{dx}}{({\mathrm{x}}^2+3+{\displaystyle \frac{1}{{\mathrm{x}}^2}})}\\ & =& \frac{1}{2}\int \frac{{\displaystyle \left(\frac{2}{{\mathrm{x}}^2}+1-1\right)\mathrm{dx}}}{({\mathrm{x}}^2+3+{\displaystyle \frac{1}{{\mathrm{x}}^2}})}\\ & =& \frac{1}{2}\int \frac{{\displaystyle \left(\frac{1}{{\mathrm{x}}^2}+1-1+\frac{1}{{\mathrm{x}}^2}\right)\mathrm{dx}}}{({\mathrm{x}}^2+3+{\displaystyle \frac{1}{{\mathrm{x}}^2}})}\\ & =& \frac{1}{2}\int \frac{{\displaystyle \left(\left(\frac{1}{{\mathrm{x}}^2}+1\right)-\left(1-\frac{1}{{\mathrm{x}}^2}\right)\right)\mathrm{dx}}}{({\mathrm{x}}^2+3+{\displaystyle \frac{1}{{\mathrm{x}}^2}})}\\ & =& \frac{1}{2}\int \frac{{\displaystyle \left(\frac{1}{{\mathrm{x}}^2}+1\right)\mathrm{dx}}}{({\mathrm{x}}^2+3+{\displaystyle \frac{1}{{\mathrm{x}}^2}})}-\frac{1}{2}\int \frac{{\displaystyle \left(1-\frac{1}{{\mathrm{x}}^2}\right)\mathrm{dx}}}{({\mathrm{x}}^2+3+{\displaystyle \frac{1}{{\mathrm{x}}^2}})}\\ & =& \frac{1}{2}\int \frac{{\displaystyle \left(1+\frac{1}{{\mathrm{x}}^2}\right)\mathrm{dx}}}{({\mathrm{x}}^2-2\left(\mathrm{x}\right)\left({\displaystyle \frac{1}{\mathrm{x}}}\right)+{\displaystyle \frac{1}{{\mathrm{x}}^2}+5})}-\frac{1}{2}\int \frac{{\displaystyle \left(1-\frac{1}{{\mathrm{x}}^2}\right)\mathrm{dx}}}{({\mathrm{x}}^2+2\left(\mathrm{x}\right)\left(\frac{1}{\mathrm{x}}\right)+{\displaystyle \frac{1}{{\mathrm{x}}^2}+1})}\\ & =& \frac{1}{2}\int \frac{{\displaystyle \left(1+\frac{1}{{\mathrm{x}}^2}\right)\mathrm{dx}}}{{\left(\mathrm{x}-{\displaystyle \frac{1}{\mathrm{x}}}\right)}^2+{\left(\sqrt{5}\right)}^2}-\frac{1}{2}\int \frac{{\displaystyle \left(1-\frac{1}{{\mathrm{x}}^2}\right)\mathrm{dx}}}{{(\mathrm{x}+{\displaystyle \frac{1}{\mathrm{x}}})}^2+{\left(1\right)}^2}\\ & =& \frac{1}{2}\int \frac{{\displaystyle \mathrm{du}}}{{\left(\mathrm{u}\right)}^2+{\left(\sqrt{5}\right)}^2}-\frac{1}{2}\int \frac{{\displaystyle \mathrm{dv}}}{{\left(\mathrm{v}\right)}^2+{\left(1\right)}^2}\left[\mathbf{Substitute}\mathbf{}\mathbf{x}\mathbf{-}\frac{\mathbf{1}}{\mathbf{x}}\mathbf{=}\mathbf{u}\mathbf{\Rightarrow }\mathbf{(}\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{)}\mathbf{dx}\mathbf{=}\mathbf{du}\mathbf{}\mathbf{andx}\mathbf{+}\frac{\mathbf{1}}{\mathbf{x}}\mathbf{=}\mathbf{v}\mathbf{\Rightarrow }\mathbf{(}\mathbf{1}\mathbf{-}\frac{\mathbf{1}}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{)}\mathbf{dx}\mathbf{=}\mathbf{d}\mathbf{v}\mathbf{}\right]\\ & =& \frac{1}{2}\left[\frac{1}{\sqrt{5}}{\tan }^{-1}\left(\frac{\mathrm{u}}{\sqrt{5}}\right)-{\tan }^{-1}\left(\mathrm{v}\right)\right]+{\mathrm{c}}_2\left[\mathbf{\because }\mathbf{\int }\frac{{\displaystyle \mathrm{dx}}}{{\mathbf{\left(}\mathbf{x}\mathbf{\right)}}^{\mathbf{2}}\mathbf{+}{\mathbf{\left(}\mathbf{a}\mathbf{\right)}}^{\mathbf{2}}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{a}}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\frac{\mathbf{x}}{\mathbf{a}}\mathbf{\right)}\right]\\ & =& \frac{1}{2}\left[\frac{1}{\sqrt{5}}{\tan }^{-1}\left(\frac{\mathrm{x}-{\displaystyle \frac{1}{\mathrm{x}}}}{\sqrt{5}}\right)-{\tan }^{-1}\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)\right]+{\mathrm{c}}_2\\ & =& \frac{1}{2\sqrt{5}}{\tan }^{-1}\left(\frac{{\mathrm{x}}^2-1}{\sqrt{5}\mathrm{x}}\right)-\frac{1}{2}{\tan }^{-1}\left(\frac{{\mathrm{x}}^2+1}{\mathrm{x}}\right)+{\mathrm{c}}_2\mathbf{\_\_\_\_\_\_\_}\mathbf{\left(}\mathbf{2}\mathbf{\right)}\mathbf{}\end{array}$

Step4. Calculate the value of $\alpha ,\beta ,\gamma ,\delta$:

$\begin{array}{rcl}\int \frac{\left(x^2-1\right)+{\tan }^{-1}\left({\displaystyle \frac{x^2+1}{x}}\right)dx}{\left(x^4+3x^2+1\right){\tan }^{-1}\left(\frac{x^2+1}{x}\right)}& =& \alpha {\log }_e\left({\tan }^{-1}\left(\frac{x^2+1}{x}\right)\right)+\beta {\tan }^{-1}\left[\gamma \left(\frac{x^2+1}{x}\right)\right]+\delta {\tan }^{-1}\left(\frac{x^2+1}{x}\right)\\ & \Rightarrow & {\log }_e\left|{\tan }^{-1}\left(x+\frac{1}{x}\right)\right|+\frac{1}{2\sqrt{5}}{\tan }^{-1}\left(\frac{{\mathrm{x}}^2-1}{\sqrt{5}\mathrm{x}}\right)-\frac{1}{2}{\tan }^{-1}\left(\frac{{\mathrm{x}}^2+1}{\mathrm{x}}\right)+C=\alpha {\log }_e\left({\tan }^{-1}\left(\frac{x^2+1}{x}\right)\right)+\beta {\tan }^{-1}\left[\gamma \left(\frac{x^2+1}{x}\right)\right]+\delta {\tan }^{-1}\left(\frac{x^2+1}{x}\right)\\ & \Rightarrow & \alpha =1\mathbf{\left[}\mathbf{Substitute}\mathbf{}\mathbf{the}\mathbf{}\mathbf{value}\mathbf{}\mathbf{from}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{}\mathbf{and}\mathbf{}\mathbf{\left(}\mathbf{2}\mathbf{\right)}\mathbf{\right]}\\ \beta & =& \frac{1}{2\sqrt{5}}\\ \gamma & =& \frac{1}{\sqrt{5}}\mathbf{[}\mathbf{Compare}\mathbf{}\mathbf{left}\mathbf{}\mathbf{hand}\mathbf{}\mathbf{side}\mathbf{}\mathbf{and}\mathbf{}\mathbf{right}\mathbf{}\mathbf{hand}\mathbf{}\mathbf{side}\mathbf{}\mathbf{]}\\ \delta & =& \frac{-1}{2}\end{array}$

Step5. Calculate the value of $10\left(\alpha +\beta \gamma +\delta \right)$

$$\begin{gathered} 10\left(\alpha +\beta \gamma +\delta \right) \\ =10\left(1+\left(\frac{1}{2\sqrt{5}}\right)\left(\frac{1}{\sqrt{5}}\right)+\left(\frac{-1}{2}\right)\right)\left[Substitutethevalueof\alpha ,\beta ,\gamma ,\delta \right] \\ =10\left(1+\frac{1}{10}-\frac{1}{2}\right) \\ =10\left(\frac{6}{10}\right) \\ =\mathbf{6} \end{gathered}$$

Hence, the value of$\mathbf{10}\left(\alpha +\mathrm{\beta \gamma }+\delta \right)\mathbf{}\mathbf{=}\mathbf{6}$