Question
$\int \frac{e^x}{x+1}[1+(x+1) \log (x+1)] d x$ equals
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Answer
$\text { (d) : Let } I=\int \frac{e^x}{x+1}[1+(x+1) \log (x+1)] d x$
$=\int e^x\left[\frac{1}{x+1}+\log (x+1)\right] d x$
It is of the form $\int e^x\left[f(x)+f^{\prime}(x) d x\right]$,
where $f(x)=\log (x+1)$ and $f^{\prime}(x)=\frac{1}{x+1}$
So, $I=e^x \log (x+1)+C$
$=\int e^x\left[\frac{1}{x+1}+\log (x+1)\right] d x$
It is of the form $\int e^x\left[f(x)+f^{\prime}(x) d x\right]$,
where $f(x)=\log (x+1)$ and $f^{\prime}(x)=\frac{1}{x+1}$
So, $I=e^x \log (x+1)+C$
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