int frac{log _e x}{left(1+log _e xright)^2} d | vedclass

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

Question

(C) Let $I = \int \frac{\log _e x}{(1+\log _e x)^2} dx$. Substitute $\log _e x = t$,which implies $x = e^t$ and $dx = e^t dt$. Substituting these into

Vedclass · Accepted Answer

$\frac{x}{\left(1+\log _e x\right)}+C$

Vedclass · Answer

(C) Let $I = \int \frac{\log _e x}{(1+\log _e x)^2} dx$. Substitute $\log _e x = t$,which implies $x = e^t$ and $dx = e^t dt$. Substituting these into the integral,we get: $I = \int \frac{t}{(1+t)^2} e^t dt$. We can rewrite the numerator as $(t+1-1)$: $I = \int \frac{t+1-1}{(1+t)^2} e^t dt = \int \left( \frac{1}{1+t} - \frac{1}{(1+t)^2} \right) e^t dt$. Using the standard integral formula $\int e^x [f(x) + f'(x)] dx = e^x f(x) + C$,where $f(t) = \frac{1}{1+t}$ and $f'(t) = -\frac{1}{(1+t)^2}$: $I = e^t \left( \frac{1}{1+t} \right) + C$. Substituting $t = \log _e x$ back into the equation: $I = e^{\log _e x} \left( \frac{1}{1+\log _e x} \right) + C = \frac{x}{1+\log _e x} + C$.

$\int \frac{\log _e x}{\left(1+\log _e x\right)^2} d x=$

Similar Questions

If $\int e^x \left( \frac{x^2-8x+19}{(x-1)^5} \right) dx = \frac{e^x(lx+m)}{(x-1)^4} + C$, then $4l+m=$

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

If $\int \frac{3-x^2}{1-2 x+x^2} e^x d x=e^x f(x)+c$,then find $f(x)$.

Integrate the function: $\frac{2+\sin 2x}{1+\cos 2x} e^x$

$\int [\sin(\log x) + \cos(\log x)] dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module