Evaluate the integral: int frac{log x}{(1+log | vedclass

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(B) Let $I = \int \frac{\log x}{(1+\log x)^2} dx$. Substitute $\log x = t$, which implies $x = e^t$ and $dx = e^t dt$. Then $I = \int \frac{t}{(1+t)^2

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $I = \int \frac{\log x}{(1+\log x)^2} dx$. Substitute $\log x = t$, which implies $x = e^t$ and $dx = e^t dt$. Then $I = \int \frac{t}{(1+t)^2} e^t dt$. Rewrite the numerator as $(t+1-1)$: $I = \int \frac{t+1-1}{(1+t)^2} e^t dt$. $I = \int \left( \frac{1}{1+t} - \frac{1}{(1+t)^2} \right) e^t dt$. 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 $f'(t) = -\frac{1}{(1+t)^2}$. Thus, $I = e^t \left( \frac{1}{1+t} \right) + c$. Substituting back $t = \log x$ and $e^t = x$: $I = \frac{x}{1+\log x} + c$.

Evaluate the integral: $\int \frac{\log x}{(1+\log x)^2} dx$

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