If $\mathrm{f}(x)=1+x ; \mathrm{g}(x)=\log x$, then $\int \mathrm{g}(\mathrm{f}(x)) \mathrm{d} x$ is equal to

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Aicharya · Accepted Answer

$\begin{aligned} & \int \mathrm{g}(\mathrm{f}(x)) \mathrm{d} x \ & =\int 1 	imes \log (1+x) \mathrm{d} x\end{aligned}$ $\begin{aligned} & =x \log (1+x)-\int x 	imes \frac{1}{(1+x)} \mathrm{d} x+\mathrm{c} \ & =x \log (1+x)-\left[\int \frac{1+x}{1+x} \mathrm{~d} x-\int \frac{1}{1+x} \mathrm{~d} xight]+\mathrm{c} \ & =x \log (1+x)-x+\log (1+x)+\mathrm{c} \ & =(1+x) \log (1+x)-x+\mathrm{c}\end{aligned}$

If $\mathrm{f}(x)=1+x ; \mathrm{g}(x)=\log x$, then $\int \mathrm{g}(\mathrm{f}(x)) \mathrm{d} x$ is equal to

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