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I want to solve this integral and need some directions. $$\int^1_0\frac{dx}{\sqrt{x+3}-1}$$ I decided to call $x+3 = t^2 \rightarrow 2tdt = dx$ then : $$\int^1_0 \frac {2tdt}{t^2-1}$$ Now what should I do? call $t^2-1 = u$ ? and do the same thing on u?
Thanks!

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  • $\begingroup$ OK, but after I do that I get this: $$\int^1_0 \frac{du}{u} \rightarrow ln|t^2-1| = ln|x+3-1| = ln|x+2| $$ and this is not the F(x) of this function. $\endgroup$ – Ofir Attia May 16 '13 at 11:08
  • $\begingroup$ You have got your expression wrong in the second expression. $\sqrt{x+3}=t$ right? Your expression is $\int\dfrac{2tdt}{t-1}$ $\endgroup$ – Inceptio May 16 '13 at 11:13
  • $\begingroup$ no I set it $t^2$ ( without the square root ). $\endgroup$ – Ofir Attia May 16 '13 at 11:15
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    $\begingroup$ I write it again: $$\int \frac{dx}{\sqrt{x+3}-1}$$ then: $t^2=x+3$ and $t=\sqrt{x+3}$ and the derivative is: $2tdt=dx$ after that we get : $$\int \frac{2tdt}{t-1}$$ now what? I can seperate them like : $$\int (2+ \frac{2}{t-1})dt$$? $\endgroup$ – Ofir Attia May 16 '13 at 11:21
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    $\begingroup$ Indeed. You could write this as an answer to your own question (presumably you know how to do the last integral). $\endgroup$ – Lord_Farin May 16 '13 at 11:42
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Hints:

Plug in $\sqrt{x+3}=t^2$, $dx=4t^3 dt$

$\int \dfrac{4t^3}{t^2-1}=\int\dfrac{2t^2}{t-1}+ \int\dfrac{2t^2}{t+1}$

Now use integration by parts for the each of the expression.

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Let $\sqrt{x+3}-1=t\implies x=(t+1)^2-3=t^2+2t-2\implies dx=2(t+1)dt$

If $x=0\implies t=\sqrt{3}-1, x=1\implies t=\sqrt{1+3}-1=1$

$$\int^1_0\frac{dx}{\sqrt{x+3}-1}=\int_{\sqrt3-1}^1\frac{2(t+1)}t dt$$

$$=2\int_{\sqrt3-1}^1\left(1+\frac1t\right) dt$$

Can you take it from here?

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