Question:
Evaluate the following integrals:
$\int \log (x+1) d x$
Solution:
Let $I=\int \log (x+1) d x$
That is,
$I=\int 1 \times \log (x+1) d x$
Using integration by parts,
$I=\log (x+1) \int 1 d x-\int \frac{d}{d x} \log (x+1) \int 1 d x$
We know that, $\int 1 \mathrm{dx}=\mathrm{x}$ and $\int \log \mathrm{x}=\frac{1}{\mathrm{x}}$
$\frac{x}{x+1}=1-\frac{1}{x+1}$
$=x \log (x+1)-\int\left(1-\frac{1}{x+1}\right) d x$
$=x \log (x+1)-x+\log (x+1)+c$
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