int log x^2 , dx = . . . . . . + C. | vedclass

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

Question

(D) We need to evaluate the integral $I = \int \log x^2 \, dx$.Using the property of logarithms,$\log x^2 = 2 \log x$.So,$I = \int 2 \log x \, dx = 2

Vedclass · Accepted Answer

$2 x \log \left(\frac{x}{e}\right)$

Vedclass · Answer

(D) We need to evaluate the integral $I = \int \log x^2 \, dx$.Using the property of logarithms,$\log x^2 = 2 \log x$.So,$I = \int 2 \log x \, dx = 2 \int \log x \, dx$.Using integration by parts,$\int \log x \, dx = x \log x - x + C_1$.Therefore,$I = 2(x \log x - x) + C = 2x \log x - 2x + C$.We can rewrite this as $2x(\log x - 1) = 2x(\log x - \log e) = 2x \log \left(\frac{x}{e}\right) + C$.Thus,the correct option is $D$.

$\int \log x^2 \, dx =$ . . . . . . $+ C$.

Similar Questions

$\int x \cos x \, dx = $

$\int \tan ^{-1}\left(\sqrt{\frac{1-x}{1+x}}\right) d x$ is equal to

Integrate the function: $x \log(2x)$

$\int (\log x)^3 x^5 dx = $

If $\int x^2 \cos^2 x \, dx = \frac{1}{6} f(x) + g(x) \sin 2x + h(x) \cos 2x + c$,then $f(1) + g(2) + h(\frac{1}{2}) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module