Question

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

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

$=\alpha {\log }_e\left({\tan }^{-1}\left(\frac{x^2+1}{x}\right)\right)$

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

where $C$ is an arbitrary constant, then the value

of $10(\alpha +\beta \gamma +\delta )$ is equal to

Answer 1

Answer: 6

$\int \frac{\left(x^2-1\right)dx}{\left(x^4+3x^2+1\right){\tan }^{-1}\left(x+\frac{1}{x}\right)}+\int \frac{dx}{x^4+3x^2+1}$
$\int \frac{\left(1-\frac{1}{x^2}\right)dx}{\left({\left(x+\frac{1}{x}\right)}^2+1\right){\tan }^{-1}\left(x+\frac{1}{x}\right)}+\frac{1}{2}\int \frac{\left(x^2+1\right)-\left(x^2-1\right)dx}{x^4+3x^2+1}$
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 $x-\frac{1}{x}=y,x+\frac{1}{x}=z$
${\log }_{e}t+\frac{1}{2}\int \frac{dy}{y^2+5}-\frac{1}{2}\int \frac{dz}{z^2+1}$
$={\log }_e{\tan }^{-1}\left(x+\frac{1}{x}\right)+\frac{1}{2\sqrt{5}}{\tan }^{-1}\left(\frac{x^2-1}{\sqrt{5}x}\right)$
$-\frac{1}{2}{\tan }^{-1}\left(\frac{x^2+1}{x}\right)+C$
$\alpha =1,\beta =\frac{1}{2\sqrt{5}},\gamma =\frac{1}{\sqrt{5}},\delta =\frac{-1}{2}$
or
$\alpha =1,\beta =\frac{-1}{2\sqrt{5}},\gamma =\frac{-1}{\sqrt{5}},\delta =\frac{-1}{2}$
$10(\alpha +\beta \gamma +\delta )=10\left(1+\frac{1}{10}-\frac{1}{2}\right)=6$