Integral Calculus
Integral Calculus
star_batch_jee_advanced_2025
Grade 12
Question:
If $\int \left[\left(\frac{x}{e}\right)^x + \left(\frac{e}{x}\right)^x\right] \ln udx = A\left(\frac{x}{e}\right)^x + B\left(\frac{e}{x}\right)^x + C$, then the value of $A + B$ is
Step-by-Step Solution
Key Concept: Recognize the substitution $t = \left(\frac{x}{e}\right)^x$ transforms the integrand into a simpler form involving $\ln t$ and its derivative.
Let $t = \left(\frac{x}{e}\right)^x$, so $x \ln\left(\frac{x}{e}\right) = \ln t$. Differentiating gives $(1 + \ln x - \ln e)dx = \frac{1}{t}dt$, which simplifies to $(\ln x)dx = \frac{1}{t}dt$. Alternatively, use $x \ln\left(\frac{x}{e}\right) = \ln t$ to get $(1 + \ln x - \ln e)dx = \frac{1}{t}dt$. The integral evaluates to $I = t + \ln t - t + C = \ln t - t + C$.
Correct Answer: 1,2,3,4