Evaluate the definite integral: int_{1}^{e} ( | vedclass

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

Question

(D) Let $I = \int_{1}^{e} (x+1) e^{x} \ln x \, dx$. We know that $\int e^{x} (f(x) + f'(x)) \, dx = e^{x} f(x) + C$. Rewrite the integrand as: $(x+1)

Vedclass · Accepted Answer

$e^{e} - 1$

Vedclass · Answer

(D) Let $I = \int_{1}^{e} (x+1) e^{x} \ln x \, dx$. We know that $\int e^{x} (f(x) + f'(x)) \, dx = e^{x} f(x) + C$. Rewrite the integrand as: $(x+1) e^{x} \ln x = e^{x} (x \ln x + \ln x) = e^{x} (x \ln x + \ln x + 1 - 1) = e^{x} ((x \ln x) + (\ln x + 1) - 1)$. This does not immediately fit the form. Let us use integration by parts. Let $u = \ln x$ and $dv = (x+1)e^{x} dx$. Then $du = \frac{1}{x} dx$ and $v = \int (x+1)e^{x} dx = x e^{x}$. Using $\int u \, dv = uv - \int v \, du$: $I = [x e^{x} \ln x]_{1}^{e} - \int_{1}^{e} x e^{x} \cdot \frac{1}{x} \, dx$ $I = [x e^{x} \ln x]_{1}^{e} - \int_{1}^{e} e^{x} \, dx$ $I = [x e^{x} \ln x - e^{x}]_{1}^{e}$ $I = (e \cdot e^{e} \cdot \ln e - e^{e}) - (1 \cdot e^{1} \cdot \ln 1 - e^{1})$ $I = (e \cdot e^{e} \cdot 1 - e^{e}) - (0 - e)$ $I = e^{e+1} - e^{e} + e = e^{e}(e-1) + e$.

Evaluate the definite integral: $\int_{1}^{e} (x+1) e^{x} \ln x \, dx$

Similar Questions

Let $f(x) = \{x\}$ denote the fractional part of a real number $x$. Then, the value of $\int_{0}^{\sqrt{3}} f(x^2) dx$ is

If $\int\limits_0^{f(x)} {{t^2}\,dt} = x \cos(\pi x)$,then find $f'(9)$.

The value of the integral $\int_{\pi/6}^{\pi/3} \left(\frac{4 - \csc^2 x}{\cos^4 x}\right) dx$ is:

Let $f_n = \int_0^{\frac{\pi}{2}} \left(\sum_{k=1}^n \sin^{k-1} x\right) \left(\sum_{k=1}^n (2k-1) \sin^{k-1} x\right) \cos x \, dx$,where $n \in N$. Then $f_{21} - f_{20}$ is equal to $...........$.

The value of the integral $\int_{-1}^1 \frac{|x+2|}{x+2} \, dx$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module