If f(x)=1+x and g(x)=log x,then int g(f(x)) , | vedclass

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(A) We are given $f(x) = 1+x$ and $g(x) = \log x$. Therefore,$g(f(x)) = \log(1+x)$. We need to evaluate the integral $\int \log(1+x) \, dx$. Using int

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$(1+x) \log (1+x)-x+c$,(where $c$ is a constant of integration)

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(A) We are given $f(x) = 1+x$ and $g(x) = \log x$. Therefore,$g(f(x)) = \log(1+x)$. We need to evaluate the integral $\int \log(1+x) \, dx$. Using integration by parts,$\int u \, dv = uv - \int v \, du$. Let $u = \log(1+x)$ and $dv = dx$. Then $du = \frac{1}{1+x} \, dx$ and $v = x$. $\int \log(1+x) \, dx = x \log(1+x) - \int \frac{x}{1+x} \, dx + c$. We can rewrite the integrand as $\frac{x}{1+x} = \frac{1+x-1}{1+x} = 1 - \frac{1}{1+x}$. So,$\int \log(1+x) \, dx = x \log(1+x) - \int (1 - \frac{1}{1+x}) \, dx + c$. $= x \log(1+x) - (x - \log(1+x)) + c$. $= x \log(1+x) - x + \log(1+x) + c$. $= (1+x) \log(1+x) - x + c$.

If $f(x)=1+x$ and $g(x)=\log x$,then $\int g(f(x)) \, dx$ is equal to

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