Question

For real numbers \(\alpha, \beta, \gamma\) and \(\delta\), if
\(\int \frac{(x^2-1)+\tan^{-1}\left(\frac{x^2+1}{x}\right)}{(x^4+3x^2+1)\tan^{-1}\left(\frac{x^2+1}{x}\right)}dx\)
\(=\alpha\log_e\left(\tan^{-1}\left(\frac{x^2+1}{x}\right)\right)+\beta\tan^{-1}\left(\frac{\gamma(x^2+1)}{x}\right)+\delta\tan^{-1}\left(\frac{x^2+1}{x}\right)+C\)
where C is an arbitrary constant, then the value of \(10(\alpha+\beta\gamma+\delta)\) is equal to ______________.

Hint:

Split the integrand into two integrals; the first gives \(\ln(\tan^{-1}u)\) with \(u=x+\frac{1}{x}\), the second is done with the \(x-\frac{1}{x}\) and \(x+\frac{1}{x}\) substitutions.

Answer 1

Correct Answer: 6

Let \(u=\frac{x^2+1}{x}=x+\frac{1}{x}\) and \(t=\tan^{-1}u\). Note \(x^4+3x^2+1=x^2(u^2+1)\).
The integrand splits as \(\frac{x^2-1}{(x^4+3x^2+1)\,t}+\frac{1}{x^4+3x^2+1}\).
First part: \(\frac{x^2-1}{x^2(u^2+1)}dx=\frac{du}{1+u^2}=dt\), so it gives \(\int\frac{dt}{t}=\log_e t\). Hence \(\alpha=1\).
Second part: \(\int\frac{dx}{x^4+3x^2+1}=\int\frac{1/x^2}{x^2+\frac{1}{x^2}+3}dx\). Write \(\frac{1}{x^2}=\frac12\left[\left(1+\frac{1}{x^2}\right)-\left(1-\frac{1}{x^2}\right)\right]\).
With \(v=x-\frac{1}{x}\): \(x^2+\frac{1}{x^2}+3=v^2+5\), and with \(u=x+\frac1x\): \(x^2+\frac{1}{x^2}+3=u^2+1\).
So the second part \(=\frac12\int\frac{dv}{v^2+5}-\frac12\int\frac{du}{u^2+1}=\frac{1}{2\sqrt5}\tan^{-1}\frac{v}{\sqrt5}-\frac12\tan^{-1}u\).
Comparing with the given form: \(\alpha=1\), \(\beta=\frac{1}{2\sqrt5}\), \(\gamma=\frac{1}{\sqrt5}\), \(\delta=-\frac12\) (the middle term is read as \(\tan^{-1}\frac{\gamma(x^2-1)}{x}\), see notes).
\(10(\alpha+\beta\gamma+\delta)=10\left(1+\frac{1}{10}-\frac12\right)=6\).
Answer: 6