(1x^2+1x^4)([3]x+1x^3)^1/23dx = -3(+1)(3x^+x^)^(+1)/()+C, x0, ,,Z, where C is the constan…

Question

$\displaystyle\int\left(\frac1{x^2}+\frac1{x^4}\right)\left(\sqrt[3]{x}+\frac1{x^3}\right)^{1/23}dx = -\dfrac{\alpha}{3(\alpha+1)}\left(3x^\beta+x^\gamma\right)^{\frac{\alpha+1}{\alpha}}+C$, $x>0$, $\alpha,\beta,\gamma\in\mathbb{Z}$, where $C$ is the constant of integration. Then $\alpha+\beta+\gamma$ is equal to \_\_\_\_\_\_\_\_\_\_\_.

Answer: 19

Accepted Answer

Let $t = x^{1/3}+x^{-3}$, so $dt = -3\left(\dfrac1{x^2}+\dfrac1{x^4}\right)dx$:

$$\int\left(\frac1{x^2}+\frac1{x^4}\right)t^{1/23}\,dx = -\frac13\int t^{1/23}\,dt = -\frac13\cdot\frac{t^{24/23}}{24/23}+C = \frac{t^{24/23}}{-72/23}+C$$

Matching the given form gives $\alpha=23,\,\beta=-1,\,\gamma=-3$, so

$$\alpha+\beta+\gamma = 23-1-3 = 19$$

Solutions · 1