int frac{log x}{(1 + log x)^2} dx = | vedclass

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

Question

(C) Let $I = \int \frac{\log x}{(1 + \log x)^2} dx$. Substitute $1 + \log x = t$,which implies $\log x = t - 1$. Differentiating both sides,we get $\f

Vedclass · Accepted Answer

$\frac{x}{1 + \log x} + c$

Vedclass · Answer

(C) Let $I = \int \frac{\log x}{(1 + \log x)^2} dx$. Substitute $1 + \log x = t$,which implies $\log x = t - 1$. Differentiating both sides,we get $\frac{1}{x} dx = dt$,so $dx = x dt$. Since $1 + \log x = t$,we have $\log x = t - 1$,so $x = e^{t-1}$. Thus,$dx = e^{t-1} dt$. Substituting these into the integral: $I = \int \frac{t - 1}{t^2} e^{t-1} dt = \int e^{t-1} \left( \frac{t}{t^2} - \frac{1}{t^2} \right) dt = \int e^{t-1} \left( \frac{1}{t} - \frac{1}{t^2} \right) dt$. Using the standard integral formula $\int e^u (f(u) + f'(u)) du = e^u f(u) + c$,where $f(t) = \frac{1}{t}$ and $f'(t) = -\frac{1}{t^2}$: $I = e^{t-1} \cdot \frac{1}{t} + c$. Substituting $t = 1 + \log x$ back: $I = \frac{e^{(1 + \log x) - 1}}{1 + \log x} + c = \frac{e^{\log x}}{1 + \log x} + c = \frac{x}{1 + \log x} + c$.

$\int \frac{\log x}{(1 + \log x)^2} dx = $

Similar Questions

$\int {{{\left( {\frac{{x + 2}}{{x + 4}}} \right)}^2}{e^x}\,dx} $ is equal to

$\int e^x \left(\frac{x+2}{x+4}\right)^2 dx =$

$\int e^{x} \frac{(x-1)}{x^{2}} d x$ is equal to

If $\int {{e^{{x^2}}}\left( {2 - \frac{1}{{{x^2}}}} \right)dx = {e^{{x^2}}}f(x) + C} $ and $f\left( {\frac{1}{2}} \right) = 2$,then $f(1)$ is equal to (where $C$ is an arbitrary constant).

If $\int e^{\alpha x}\left(\frac{1-\beta \sin x}{1-\cos x}\right) d x=-e^x \cot \frac{x}{2}+c$, then $\frac{\alpha^2+\beta^2}{2 \alpha \beta}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module