Question

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

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

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

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

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

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

Answer 1

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