For real numbers ,,,, if [(x^2-1)+^-1(x^2+1x)(x^4+3x^2+1)^-1(x^2+1x),dx = _e|^-1(x^2+1x)|+…

Question

For real numbers $\alpha,\beta,\gamma,\delta$, 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..........

Answer: 6

Accepted Answer

Split the integral as $I_1+I_2$ where $I_1=\displaystyle\int\dfrac{x^2-1}{(x^4+3x^2+1)\tan^{-1}\left(\frac{x^2+1}x\right)}dx$ and $I_2=\displaystyle\int\dfrac{dx}{x^4+3x^2+1}$.

For $I_1$: let $t=\tan^{-1}\left(\dfrac{x^2+1}x\right)$; one verifies $dt = \dfrac{x^2-1}{x^4+3x^2+1}\,dx$, so $I_1=\ln|t|+C_1 = \log_e\left|\tan^{-1}\left(\dfrac{x^2+1}x\right)\right|+C_1$.

For $I_2$: write $\dfrac1{x^4+3x^2+1}=\dfrac12\left[\dfrac{x^2+1}{x^4+3x^2+1}-\dfrac{x^2-1}{x^4+3x^2+1}\right]$, and dividing by $x^2$ throughout, each piece is a standard $\tan^{-1}$ form:

$$I_2 = \frac1{2\sqrt5}\tan^{-1}\left(\frac{x^2-1}{\sqrt5\,x}\right)-\frac12\tan^{-1}\left(\frac{x^2+1}x\right)+C_2$$

Comparing: $\alpha=1$, $\beta=\dfrac1{2\sqrt5}$, $\gamma=\dfrac1{\sqrt5}$, $\delta=-\dfrac12$.

$$10(\alpha+\beta\gamma+\delta) = 10\left(1+\frac1{10}-\frac12\right) = 6$$

Solutions · 1