Indefinite Integration
Power Function Integral — Matching Coefficients
nta_pyq_2023_apr
Grade None
Question:
For $\alpha,\beta,\gamma,\delta\in\mathbb{N}$, if $\displaystyle\int\!\left[\left(\frac{x}{e}\right)^{2x}+\left(\frac{e}{x}\right)^{2x}\right]\ln x\,dx=\dfrac{1}{\alpha}\!\left(\frac{x}{e}\right)^{\beta x}-\dfrac{1}{\gamma}\!\left(\frac{e}{x}\right)^{\delta x}+C$, then $\alpha+2\beta+3\gamma-4\delta$ is equal to
Step-by-Step Solution
Key Concept: Let $t=(\frac{x}{e})^{2x}$. Then $2x(\ln x-1)=\ln t\Rightarrow\ln x\,dx=\frac{dt}{2t}$. So $\int((\frac{x}{e})^{2x}+(\frac{e}{x})^{2x})\ln x\,dx=\frac{1}{2}\int(1+t^{-2})dt$.
$\alpha=\beta=\gamma=\delta=2$. $\alpha+2\beta+3\gamma-4\delta=4$.
Correct Answer: 2