Integration by parts.

How can I integrate $\int_{3}^8 \ln \sqrt{x+1}\ dx$ by parts ?

Is this step right ?

$\int_{3}^8 \frac{1}{2}\ln(x+1)\ dx$ = $\frac{1}{2} \int_{3}^8\ln(x+1)\ dx$

$f^{'}(x) = 1 , f(x) = x , g(x)= ln(x+1)$

$\left[ x\ln(x+1) \right]_{x=3}^{x=8} - \int_{3}^8 x\frac{1}{x+1}\ dx$

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$f(x)=x+1$ is easier. – i707107 Feb 10 '14 at 21:06

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.

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But why the answer in the book is $9ln3 - 4ln2 - \frac{5}{2}$ ? When we plug 8 and 3 in the final equation we have $[(9)ln(9)-8] - [(4)ln(4)-3]$ which is different from the answer of the book. – Out Of Bounds Feb 11 '14 at 2:36
Don't forget the factor of $\frac 12$. And remember the logarithm laws: If we expand, we get $$\frac 92\ln(9) - \frac 12(4)\ln(4) - \frac 12(8) + \frac 12(3) \\ = 9 \ln(9^{1/2}) - 2\ln (2^2) - \frac 52 \\ = 9\ln(3) - 4\ln(2) - \frac 52$$ – amWhy Feb 11 '14 at 12:49

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 $\frac{x}{x+1} = 1-\frac{1}{x+1}.$

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But the answer in the book is $9 ln3 - 4 ln2- \frac{5}{2}$ not $4ln9-\frac{3}{2}ln4$ – Out Of Bounds Feb 10 '14 at 21:32
My computations were not finished here. In any case, integration by parts and then writing $\frac{x}{x+1}=1-\frac{1}{x+1}$ is the way to go. – Martingalo Feb 10 '14 at 21:36

$\int_{3}^8 ln\ \sqrt{x+1}\ dx$= $1/2\int_{3}^8 ln\ ({x+1})\ dx$= $x.ln(x+1)|_{3}^8 + \int_{3}^8 x/(x+1) dx$ = $x.ln(x+1)|_{3}^8 + \int_{3}^8 (1- (1/1+x)) dx$ . I think from here on it"s pretty obvious.

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But ... the quetion is: how to do it by parts... – GEdgar Feb 10 '14 at 22:00