Integration by parts. How can I integrate $ \int_{3}^8 \ln \sqrt{x+1}\ dx$ by parts ?
Is this step right ?
$  \int_{3}^8 \frac{1}{2}\ln(x+1)\ dx $ = $  \frac{1}{2} \int_{3}^8\ln(x+1)\ dx$
$f^{'}(x) = 1 , f(x) = x , g(x)= ln(x+1) $
$ \left[ x\ln(x+1) \right]_{x=3}^{x=8} - \int_{3}^8 x\frac{1}{x+1}\ dx  $
 A: Note:
There is/was nothing stopping you from taking $f'(x) = 1$ and $f(x) = x+1$.
You can then clean up your last integral very nicely, with numerator and denominator canceling.
You'd have: $$\int_{3}^8 \frac{1}{2}\ \ln(x+1)\ dx=\Big[ \frac 12(x+1) \ln(x+1) \Big]_{x=3}^{x=8}\;\; - \;\;\frac 12\int_{3}^8\, dx \;\; =\frac 12 \Big[ (x+1) \ln(x+1) - x\Big]_{x=3}^{x=8}$$
Either route ultimately yields the same result.
A: Everything you wrote is right.
Now take the derivative of $g(x)=\ln (x+1)$. So $g'(x)=\frac{1}{x+1}$. Then by the integration by parts formula you have
$$\frac{1}{2}\int_3^8 \ln(x+1)dx= \frac{1}{2}\left( x\ln(x+1) \bigg|_{3}^8 - \int_3^8 \frac{x}{x+1}dx\right).$$
Altogether
$$\frac{1}{2}\int_3^8 \ln(x+1)dx=  4\ln (9)-\frac{3}{2} \ln 4 - \frac{1}{2}\int_3^8 \frac{x}{x+1}dx$$
Finally the last integral can be solved by writing $\frac{x}{x+1} = 1-\frac{1}{x+1}.$
A: $\int_{3}^8 ln\ \sqrt{x+1}\ dx $= $ 1/2\int_{3}^8 ln\ ({x+1})\ dx $= $x.ln(x+1)|_{3}^8  + \int_{3}^8 x/(x+1) dx$ = $x.ln(x+1)|_{3}^8  + \int_{3}^8 (1- (1/1+x)) dx$ . I think from here on it"s pretty obvious.
