Mathematics - Indefinite Integration Question with Solution | TestHub

$\int \frac{\left(x^2-1\right)dx}{\left(x^4+3x^2+1\right){\tan }^{-1}\left(x+\frac{1}{x}\right)}+\int \frac{dx}{x^4+3x^2+1}$

$\int \frac{\left(1-\frac{1}{x^2}\right)dx}{\left({\left(x+\frac{1}{x}\right)}^2+1\right){\tan }^{-1}\left(x+\frac{1}{x}\right)}+\frac{1}{2}\int \frac{\left(x^2+1\right)-\left(x^2-1\right)dx}{x^4+3x^2+1}$

Put ${\tan }^{-1}\left(x+\frac{1}{x}\right)=t$

$\int \frac{dt}{t}+\frac{1}{2}\int \frac{\left(1+\frac{1}{x^2}\right)dx}{{\left(x-\frac{1}{x}\right)}^2+5}-\frac{1}{2}\int \frac{\left(1-\frac{1}{x^2}\right)dx}{{\left(x+\frac{1}{x}\right)}^2+1}$

Put $x-\frac{1}{x}=y,x+\frac{1}{x}=z$

${\log }_{e}t+\frac{1}{2}\int \frac{dy}{y^2+5}-\frac{1}{2}\int \frac{dz}{z^2+1}$

$={\log }_{\text{e}}{\tan }^{-1}\left(x+\frac{1}{x}\right)+\frac{1}{2\sqrt{5}}{\tan }^{-1}\left(\frac{x^2-1}{\sqrt{5}x}\right)$

$-\frac{1}{2}{\tan }^{-1}\left(\frac{x^2+1}{x}\right)+C$

$\alpha =1,\beta =\frac{1}{2\sqrt{5}},\gamma =\frac{1}{\sqrt{5}},\delta =\frac{-1}{2}$

or $\alpha =1,\beta =\frac{-1}{2\sqrt{5}},\gamma =\frac{-1}{\sqrt{5}},\delta =\frac{-1}{2}$

$10(\alpha +\beta \gamma +\delta )=10\left(1+\frac{1}{10}-\frac{1}{2}\right)=6$