int (2sin x + frac{1}{x}) , dx is equal to | vedclass

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

Question

(A) To evaluate the integral $\int (2\sin x + \frac{1}{x}) \, dx$,we use the linearity property of integration:$\int (2\sin x + \frac{1}{x}) \, dx = \

Vedclass · Accepted Answer

$-2\cos x + \log |x| + c$

Vedclass · Answer

(A) To evaluate the integral $\int (2\sin x + \frac{1}{x}) \, dx$,we use the linearity property of integration:$\int (2\sin x + \frac{1}{x}) \, dx = \int 2\sin x \, dx + \int \frac{1}{x} \, dx$We know that $\int \sin x \, dx = -\cos x + c_1$ and $\int \frac{1}{x} \, dx = \log |x| + c_2$.Therefore,$\int 2\sin x \, dx + \int \frac{1}{x} \, dx = 2(-\cos x) + \log |x| + c = -2\cos x + \log |x| + c$.Thus,the correct option is $A$.

$\int (2\sin x + \frac{1}{x}) \, dx$ is equal to

Similar Questions

$\int \left( \frac{1}{x^2} + \frac{\sin^3 x + \cos^3 x}{\sin^2 x \cos^2 x} \right) dx =$

$\int \frac{dx}{\cos 2x - \cos^2 x} = $

$\int \frac{dx}{\sqrt{x + a} + \sqrt{x + b}} = $

If $\int \frac{dx}{\sqrt{16-9x^2}} = A \sin^{-1}(Bx) + C$,then $A+B=$

The value of $\int \frac{1}{x^4} \, dx$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module