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Question: For real numbers $\alpha, \beta, \gamma$ and $\delta$, if $\int \frac{(x^2 - 1) + tan^{-1} (\frac{x...

For real numbers α,β,γ\alpha, \beta, \gamma and δ\delta, if

∫(x2−1)+tan−1(x2+1x)(x4+3x2+1)tan−1(x2+1x)dx=αloge(tan−1(x2+1x))+βtan−1(γ(x2−1)x)+δtan−1(x2+1x)+C\int \frac{(x^2 - 1) + tan^{-1} (\frac{x^2+1}{x})}{(x^4 + 3x^2 + 1)tan^{-1}(\frac{x^2+1}{x})} dx = \alpha log_e (tan^{-1} (\frac{x^2+1}{x})) + \beta tan^{-1} (\frac{\gamma(x^2-1)}{x}) + \delta tan^{-1} (\frac{x^2+1}{x}) + C

where C is an arbitrary constant, then the value of 10(α+βγ+δ)10(\alpha + \beta\gamma + \delta) is equal to

Answer

6

Explanation

Solution

The integral is split into two parts by separating the numerator. The first part, ∫1tan⁡−1(x2+1x)⋅x2−1x4+3x2+1dx\int \frac{1}{\tan^{-1}(\frac{x^2+1}{x})} \cdot \frac{x^2 - 1}{x^4 + 3x^2 + 1} dx, is solved by substituting u=tan⁡−1(x2+1x)u = \tan^{-1}(\frac{x^2+1}{x}). This yields du=x2−1x4+3x2+1dxdu = \frac{x^2 - 1}{x^4 + 3x^2 + 1} dx, transforming the integral into ∫1udu=log⁡e∣u∣\int \frac{1}{u} du = \log_e |u|. This identifies α=1\alpha=1.

The second part, ∫1x4+3x2+1dx\int \frac{1}{x^4 + 3x^2 + 1} dx, is solved by dividing the numerator and denominator by x2x^2, then splitting the numerator 1x2\frac{1}{x^2} into 12((1+1x2)−(1−1x2))\frac{1}{2}\left(\left(1+\frac{1}{x^2}\right) - \left(1-\frac{1}{x^2}\right)\right). This creates two new integrals:

  1. 12∫1+1x2x2+3+1x2dx\frac{1}{2} \int \frac{1+\frac{1}{x^2}}{x^2+3+\frac{1}{x^2}} dx: Substitute p=x−1xp = x - \frac{1}{x}, so dp=(1+1x2)dxdp = (1+\frac{1}{x^2})dx and x2+3+1x2=p2+5x^2+3+\frac{1}{x^2} = p^2+5. This integral becomes 12∫dpp2+5=125tan⁡−1(p5)=125tan⁡−1(x2−1x5)\frac{1}{2} \int \frac{dp}{p^2+5} = \frac{1}{2\sqrt{5}} \tan^{-1}\left(\frac{p}{\sqrt{5}}\right) = \frac{1}{2\sqrt{5}} \tan^{-1}\left(\frac{x^2-1}{x\sqrt{5}}\right).

  2. −12∫1−1x2x2+3+1x2dx-\frac{1}{2} \int \frac{1-\frac{1}{x^2}}{x^2+3+\frac{1}{x^2}} dx: Substitute q=x+1xq = x + \frac{1}{x}, so dq=(1−1x2)dxdq = (1-\frac{1}{x^2})dx and x2+3+1x2=q2+1x^2+3+\frac{1}{x^2} = q^2+1. This integral becomes −12∫dqq2+1=−12tan⁡−1(q)=−12tan⁡−1(x2+1x)-\frac{1}{2} \int \frac{dq}{q^2+1} = -\frac{1}{2} \tan^{-1}(q) = -\frac{1}{2} \tan^{-1}\left(\frac{x^2+1}{x}\right).

Comparing the combined result with the given form of the integral yields α=1\alpha=1, β=125\beta=\frac{1}{2\sqrt{5}}, γ=15\gamma=\frac{1}{\sqrt{5}}, and δ=−12\delta=-\frac{1}{2}. Finally, calculate 10(α+βγ+δ)=10(1+(125)(15)−12)=10(1+110−12)=10(10+1−510)=10(610)=610(\alpha + \beta\gamma + \delta) = 10\left(1 + \left(\frac{1}{2\sqrt{5}}\right)\left(\frac{1}{\sqrt{5}}\right) - \frac{1}{2}\right) = 10\left(1 + \frac{1}{10} - \frac{1}{2}\right) = 10\left(\frac{10+1-5}{10}\right) = 10\left(\frac{6}{10}\right) = 6.