int_1^e {frac{{{e^x}}}{x}(1 + xlog x),dx} = | vedclass

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

Question

(A) We are given the integral $I = \int_1^e \frac{e^x}{x}(1 + x \log x) \, dx$. Expanding the integrand,we get: $I = \int_1^e \frac{e^x}{x} \, dx + \i

Vedclass · Accepted Answer

$e^e$

Vedclass · Answer

(A) We are given the integral $I = \int_1^e \frac{e^x}{x}(1 + x \log x) \, dx$. Expanding the integrand,we get: $I = \int_1^e \frac{e^x}{x} \, dx + \int_1^e e^x \log x \, dx$. Using integration by parts on the second term $\int_1^e e^x \log x \, dx$,let $u = \log x$ and $dv = e^x \, dx$. Then $du = \frac{1}{x} \, dx$ and $v = e^x$. Applying the formula $\int u \, dv = uv - \int v \, du$: $\int_1^e e^x \log x \, dx = [e^x \log x]_1^e - \int_1^e \frac{e^x}{x} \, dx$. Substituting this back into the expression for $I$: $I = \int_1^e \frac{e^x}{x} \, dx + ([e^x \log x]_1^e - \int_1^e \frac{e^x}{x} \, dx)$. The terms $\int_1^e \frac{e^x}{x} \, dx$ cancel out. $I = [e^x \log x]_1^e = (e^e \log e - e^1 \log 1)$. Since $\log e = 1$ and $\log 1 = 0$,we have: $I = e^e(1) - e(0) = e^e$.

$\int_1^e {\frac{{{e^x}}}{x}(1 + x\log x)\,dx} = $

Similar Questions

$\int {{e^x}(1 + \tan x)\sec x\,dx = }$

$\int e^{\tan^{-1} x} \left( \frac{1+x+x^2}{1+x^2} \right) dx = \rule{1cm}{0.15mm} + C$

$\int {{e^x}\sin x(\sin x + 2\cos x)} \,dx = $

The value of $\int \frac{e^{\tan ^{-1}(x)}}{1+x^2} \left[\left(\sec ^{-1} \sqrt{1+x^2}\right)^2+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right] d x$,for $x>0$ is

The integral $\int_{1}^{2} e^{x} \cdot x^{x}(1 + \log_{e} x + 1) dx$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module