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 January 28th, 2013, 12:52 PM #1 Newbie   Joined: Jan 2013 Posts: 12 Thanks: 0 Integrate rational fraction expression Integrate $\int \frac{1+\sqrt{x+1}}{1-sqrt{x+1}}\hspace{6} ,\hspace{6} x>-1$ by using substitution $t= \sqrt{x+1}$ Please help, I didn't get anything good from trying myself
 January 28th, 2013, 01:17 PM #2 Senior Member   Joined: Jul 2011 Posts: 227 Thanks: 0 Re: Integrate rational fraction expression $\int \frac{1+\sqrt{x+1}}{1-\sqrt{x+1}}dx= \int \frac{1-\sqrt{x+1}+2\sqrt{x+1}}{1-\sqrt{x+1}}dx = \int dx + 2 \int \frac{\sqrt{x+1}}{1-\sqrt{x+1}}dx$ Make the substition $1-\sqrt{x+1}=t$.
January 28th, 2013, 09:57 PM   #3
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Re: Integrate rational fraction expression

Quote:
 Originally Posted by Stud1 $\int \frac{1+\sqrt{x+1}}{1-sqrt{x+1}}dx\hspace{6} ,\hspace{6} x>-1$ by using substitution $t=\sqrt{x+1}$
[color=#FF0000]First the condition x>-1 is not correct since the function is not continuous at x=0 so that must be for x>0.[/color]

Let $t=\sqrt{x+1}\hspace{6}\Rightarrow \hspace{6}\,dt=\frac{1}{2\sqrt{x+1}}\,dx=\frac{1}{ 2t}\, dx\hspace{6}\Rightarrow \hspace{6} 2t\,dt=dx$

$\forall \,\, t>1$

So the integrand becomes :

$\int \frac{2t(1+t)}{1-t}\, dt= 2\int \,\frac{t^2+t}{1-t}dt$

Now can you perform long division ?

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