int e^{2x + log 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 $I = \int e^{2x + \log x} dx$. Using the property of exponents $e^{a+b} = e^a \cdot e^b$ and $e^{\log x} = x$,we get: $I

Vedclass · Accepted Answer

$\frac{1}{4}(2x - 1)e^{2x} + c$

Vedclass · Answer

(A) We are given the integral $I = \int e^{2x + \log x} dx$. Using the property of exponents $e^{a+b} = e^a \cdot e^b$ and $e^{\log x} = x$,we get: $I = \int e^{2x} \cdot e^{\log x} dx = \int x e^{2x} dx$. Now,apply integration by parts: $\int u v dx = u \int v dx - \int (u' \int v dx) dx$. Let $u = x$ and $v = e^{2x}$. Then $u' = 1$ and $\int v dx = \frac{e^{2x}}{2}$. $I = x \cdot \frac{e^{2x}}{2} - \int 1 \cdot \frac{e^{2x}}{2} dx$. $I = \frac{x e^{2x}}{2} - \frac{1}{2} \cdot \frac{e^{2x}}{2} + c$. $I = \frac{2x e^{2x} - e^{2x}}{4} + c = \frac{e^{2x}}{4}(2x - 1) + c$.

$\int e^{2x + \log x} dx = $

Similar Questions

Find $\int \frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}} d x$

Evaluate the integral: $\int \frac{\sqrt{x^2+1}\left[\log \left(x^2+1\right)-2 \log x\right]}{x^4} d x$

$\int e^{-2x} \sin 3x \, dx = $

$\int \sin^{-1}\left(\sqrt{\frac{x}{a+x}}\right) dx =$

$\int x{e^x} dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module