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I want to solve this integral and think about call $\sqrt{x+1} = t \rightarrow t^2 = x+1$ $$\int^3_0 \frac{dx}{1+\sqrt{x+1}}$$ Now the integral is : $$\int^3_0 \frac{2tdt}{1+t}$$ now I need your suggestions.
Thanks.

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    $\begingroup$ Substitute $1+t^2=u$ $\endgroup$ – lab bhattacharjee May 14 '13 at 7:17
  • $\begingroup$ thanks so its double substitution. $\endgroup$ – Ofir Attia May 14 '13 at 7:35
  • $\begingroup$ upper limit will be 2 in last line $\endgroup$ – iostream007 May 14 '13 at 7:48
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The substitution is a good idea. The details are not quite right. We get $dx=2t\,dt$, and therefore our definite integral is equal to $$\int_{t=1}^2 \frac{2t}{1+t}\,dt.$$ Now can you take over? It may be useful to note that $\frac{2t}{1+t}=2-\frac{2}{1+t}$.

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  • $\begingroup$ what about $t^2$? $\endgroup$ – Ofir Attia May 14 '13 at 7:31
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$$\begin{align} \int^2_0 \frac{2t\, dt}{1+t} &= 2\int^2_0 \frac{1+t-1}{1+t}\,dt \\ &= 2\int^2_0 \left( \frac{1+t}{1+t}-\frac{1}{1+t}\right)\,dt \\ &= 2\int^2_0\left( 1-\frac{1}{1+t}\right)\,dt \\ &= 2[ t-\log({1+t})]_0^2 \\ &= 2[ 2-\log 3] \\ \end{align}$$

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