Write an anti-derivative for the following fu | vedclass

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

Question

(A) We know that the derivative of $\log x$ with respect to $x$ is $\frac{1}{x}$ for $x > 0$. Also,for $x < 0$,let $y = -x$,then $\frac{d}{dx}(\log(-x

Vedclass · Accepted Answer

$\log |x|$

Vedclass · Answer

(A) We know that the derivative of $\log x$ with respect to $x$ is $\frac{1}{x}$ for $x > 0$. Also,for $x Combining these two cases,we get $\frac{d}{dx}(\log |x|) = \frac{1}{x}$ for all $x 
eq 0$. By the definition of an anti-derivative,if $\frac{d}{dx}(F(x)) = f(x)$,then $F(x)$ is an anti-derivative of $f(x)$. Therefore,$\log |x|$ is an anti-derivative of $\frac{1}{x}$.

Write an anti-derivative for the following function using the method of inspection: $\frac{1}{x}, x \neq 0$

Similar Questions

$\int \sin 2x \cos 3x \, dx = $

Find the integral of the function $\cos 2x \cos 4x \cos 6x$.

If $f^{\prime}(x)=k(\cos x+\sin x)$ and $f(0)=9, f\left(\frac{\pi}{2}\right)=15$,then $f(x)=$

$\int {\frac{{{x^3} - x - 2}}{{(1 - {x^2})}}\,dx} = $

$\int \frac{\sin 7 x}{\cos 9 x \cos 2 x} \,d x$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module