+0  
 
+15
348
4
avatar+1824 

PLEASE SHOW ALL WORKING (KEYWORD IS ALL)

EighthMersennePrime  Feb 19, 2017

Best Answer 

 #3
avatar+26752 
+15

Here's a similar derivation, but makes explicit use of partial fractions:

 

.

Alan  Feb 19, 2017
 #1
avatar+92805 
+15

ok

\(x=u^2+1\\ dx=2u\;\;du\\~\\ x-1=u^2\qquad u^2\ge0\;\;so\;\; x\ge1 \\ u=\pm\sqrt{x-1}\;\; \\ \text {Rather than having }x=u^2+1\\\text{I am going to have }u=+\sqrt{x-1 }\\ \text{Everything else is the same but I don't have to worry about the negative posibility}\\ \)

 

 

 

\(\displaystyle \int \frac{1}{2x\sqrt{x+1}}\;dx\\ =\displaystyle \int \frac{1}{2(u^2-1)u}\;2u\;\;du\\ =\displaystyle \int \frac{1}{u^2-1}\;\;du\\ =-\displaystyle \int \frac{1}{1-u^2}\;\;du\\ =-tanh^{-1}(u)+c\\ \qquad\text{equivalent for restricted values (from Wofram|alpha) }\\ \)

 

\(=\frac{1}{2} \left[ ln\frac{1-u}{1+u} \right]+c\\ =\frac{1}{2} \left[ ln\frac{1-u}{1+u} \right]+c\\ =\frac{1}{2} \left[ ln\frac{u-1}{u+1} \right]+c\\ \qquad \text{ Now c can be replaced with } logA\\ \qquad \text{Where A is a constant >0}\\ \qquad \text{I do not know why this condition was not given...}\\ =ln \left[ \frac{u-1}{u+1} \right]^{1/2}+lnA\\ =ln\left[A \left[ \frac{u-1}{u+1} \right]^{1/2}\right]\\ =ln\left[A\sqrt{ \left[ \frac{u-1}{u+1} \right]}\;\right]\\ \qquad u=\sqrt{x+1}\\ =ln\left[\;A\sqrt{ \frac{\sqrt{x+1}-1}{\sqrt{x+1}+1} }\;\;\;\right]\\ \)

 

 

 

 

 

 

 

 

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Melody  Feb 19, 2017
 #3
avatar+26752 
+15
Best Answer

Here's a similar derivation, but makes explicit use of partial fractions:

 

.

Alan  Feb 19, 2017
 #4
avatar+92805 
+10

Thanks for showing us this way Alan :)

I saw that difference of 2 squares there I thought of partial fractions but I don't know how to use them very well.  sad

I can learn from you answer :))

Melody  Feb 19, 2017

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