For ,,,N, if (x^e^2x+(e^x)/(2x))^!_e x,dx = 1 x^e^ x-1e^x x+C (reconstructed form), then +…

Question

For $\alpha,\beta,\gamma,\delta\in\mathbb{N}$, if $\displaystyle\int\left(x^{e^{2x}}+\frac{e^x}{2x}\right)^{\!}\log_e x\,dx = \dfrac1\alpha x^{e^{\beta x}}-\dfrac1\gamma\dfrac{e^x}{\delta x}+C$ (reconstructed form), then $\alpha+2\beta+3\gamma-4\delta$ is equal to:

Accepted Answer

Write $x=e^{\ln x}$, so the integrand becomes $\left[e^{2(x\ln x-x)}+e^{-2(x\ln x-x)}\right]\ln x$. Let $t=x\ln x-x$, $dt=\ln x\,dx$:

$$\int\left(e^{2t}+e^{-2t}\right)dt = \frac{e^{2t}}2-\frac{e^{-2t}}2+C = \frac12\frac{x^{2x}}{e^{2x}}-\frac12\frac{e^{2x}}{x^{2x}}+C$$

Matching the stated form gives $\alpha=\beta=\gamma=\delta=2$, so

$$\alpha+2\beta+3\gamma-4\delta = 2+4+6-8 = 4$$

Solutions · 1