# For real numbers α, β, γ and

Question:

For 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 $\mathrm{C}$ is an arbitrary constant, then the value of $10(\alpha+\beta \gamma+\delta)$ is equal to________.

Solution:

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

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

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

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

Put $x-\frac{1}{x}=y, x+\frac{1}{x}=z$

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

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

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

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

Or

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

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