int frac{log sqrt{x}}{3 x} dx is equal to | vedclass

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

Question

(A) Let $I = \int \frac{\log \sqrt{x}}{3 x} dx$. Using the property $\log \sqrt{x} = \log(x^{1/2}) = \frac{1}{2} \log x$, the integral becomes: $I = \

Vedclass · Accepted Answer

$\frac{1}{3}(\log \sqrt{x})^2+C$

Vedclass · Answer

(A) Let $I = \int \frac{\log \sqrt{x}}{3 x} dx$. Using the property $\log \sqrt{x} = \log(x^{1/2}) = \frac{1}{2} \log x$, the integral becomes: $I = \int \frac{\frac{1}{2} \log x}{3 x} dx = \frac{1}{6} \int \frac{\log x}{x} dx$. Let $u = \log x$, then $du = \frac{1}{x} dx$. Substituting these into the integral: $I = \frac{1}{6} \int u du = \frac{1}{6} \cdot \frac{u^2}{2} + C = \frac{u^2}{12} + C$. Substituting $u = \log x$ back: $I = \frac{(\log x)^2}{12} + C$. Alternatively, using the substitution $t = \log \sqrt{x} = \frac{1}{2} \log x$, then $dt = \frac{1}{2x} dx$, which implies $\frac{1}{x} dx = 2 dt$. $I = \int \frac{t}{3} (2 dt) = \frac{2}{3} \int t dt = \frac{2}{3} \cdot \frac{t^2}{2} + C = \frac{t^2}{3} + C = \frac{1}{3} (\log \sqrt{x})^2 + C$. Thus, the correct option is $A$.

Column-$I$	Column-$II$
$A$. Fishes	$(i)$. Three-chambered heart
$B$. Amphibians	$(ii)$. Incomplete double circulation
$C$. Birds	$(iii)$. Four-chambered heart
	$(iv)$. Single circulation
	$(v)$. Two-chambered heart
	$(vi)$. Double circulation

$\int \frac{\log \sqrt{x}}{3 x} dx$ is equal to

Similar Questions

The coefficient of apparent expansion of a liquid when determined using two different vessels $A$ and $B$ are $\gamma_1$ and $\gamma_2$ respectively. If the coefficient of linear expansion of the vessel $A$ is $\alpha$,then find the coefficient of linear expansion of vessel $B$.

The true solution set of the inequality,$\sqrt{5x - 6 - x^2} + \frac{\pi}{2} \int_{0}^{x} dz > x \int_{0}^{\pi} \sin^2 x \, dx$ is :

The term that accounts for intermolecular force in Vander Waals equation for non-ideal gas is:

In a transistor,if $\frac{I_C}{I_E} = \alpha$ and $\frac{I_C}{I_B} = \beta$. If $\alpha$ varies between $\frac{20}{21}$ and $\frac{100}{101}$,then the value of $\beta$ lies between:

Vedclass Test Series

Exam Paper Generator

Online Exam Module