For alpha, beta, gamma, delta in mathbb{N},if | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) We have the integral $I = \int \left( \left( \frac{x}{e} \right)^{2x} + \left( \frac{e}{x} \right)^{2x} \right) \ln x \, dx$. Note that $\left( \f

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) We have the integral $I = \int \left( \left( \frac{x}{e} \right)^{2x} + \left( \frac{e}{x} \right)^{2x} \right) \ln x \, dx$. Note that $\left( \frac{x}{e} \right)^{2x} = e^{2x \ln(x/e)} = e^{2x(\ln x - 1)} = e^{2(x \ln x - x)}$. Similarly,$\left( \frac{e}{x} \right)^{2x} = e^{-2(x \ln x - x)}$. Let $t = x \ln x - x$. Then $dt = (\ln x + x \cdot \frac{1}{x} - 1) \, dx = \ln x \, dx$. Substituting these into the integral,we get $I = \int (e^{2t} + e^{-2t}) \, dt$. Integrating,we obtain $I = \frac{e^{2t}}{2} - \frac{e^{-2t}}{2} + C$. Substituting back $t = x \ln x - x$,we get $I = \frac{1}{2} \left( \frac{x}{e} \right)^{2x} - \frac{1}{2} \left( \frac{e}{x} \right)^{2x} + C$. Comparing this with the given form,we have $\alpha = 2, \beta = 2, \gamma = 2, \delta = 2$. Thus,$\alpha + 2\beta + 3\gamma - 4\delta = 2 + 2(2) + 3(2) - 4(2) = 2 + 4 + 6 - 8 = 4$.

Similar Questions

If $n$ is a positive integer greater than $1$ and $I_{n}=\int \frac{\sin n x}{\sin x} d x$, then $I_{n+1}-I_{n-1}=$

If $I_n = \int \frac{\sin nx}{\sin x} dx$ for $n = 1, 2, 3, \ldots$,then $I_6 =$

$\int \frac{dx}{x^2 \sqrt{4+x^2}}$ is equal to

$\int \frac{x}{\sqrt{x^2-2x+5}} dx=$

Evaluate: $\int \frac{2 \cos x+1}{(2+\cos x)^2} d x - \frac{\sin x}{2+\cos x}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module