Integrate the function frac{(1+log x)^{2}}{x} | vedclass

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Question

(A) Let $1+\log x = t$. Differentiating both sides with respect to $x$,we get $\frac{1}{x} dx = dt$. Substituting these into the integral,we have: $\i

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $1+\log x = t$. Differentiating both sides with respect to $x$,we get $\frac{1}{x} dx = dt$. Substituting these into the integral,we have: $\int \frac{(1+\log x)^{2}}{x} dx = \int t^{2} dt$. Using the power rule for integration $\int x^n dx = \frac{x^{n+1}}{n+1} + C$,we get: $\int t^{2} dt = \frac{t^{3}}{3} + C$. Substituting $t = 1+\log x$ back into the expression,we get: $\frac{(1+\log x)^{3}}{3} + C$,where $C$ is an arbitrary constant.

Integrate the function $\frac{(1+\log x)^{2}}{x}$.

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