Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $

share|improve this question
4  
$f(x)=x+1$ is easier. –  i707107 Feb 10 at 21:06
add comment

3 Answers

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.

share|improve this answer
    
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. –  Tennisman Feb 11 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 at 12:49
add comment

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

share|improve this answer
    
But the answer in the book is $ 9 ln3 - 4 ln2- \frac{5}{2} $ not $ 4ln9-\frac{3}{2}ln4 $ –  Tennisman Feb 10 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 at 21:36
add comment

$\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.

share|improve this answer
    
But ... the quetion is: how to do it by parts... –  GEdgar Feb 10 at 22:00
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.