Integral of fractional expression $\int^3_0 \frac{dx}{1+\sqrt{x+1}}$

Question

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.

Substitute $1+t^2=u$ – lab bhattacharjee May 14 '13 at 07:17 — lab bhattacharjee, May 14 '13 at 07:17
thanks so its double substitution. – Ofir Attia May 14 '13 at 07:35 — Ofir Attia, May 14 '13 at 07:35
upper limit will be 2 in last line – iostream007 May 14 '13 at 07:48 — iostream007, May 14 '13 at 07:48

score 5 · Accepted Answer · answered May 14 '13 at 07:18

5

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 $\frac{2t}{1+t}=2-\frac{2}{1+t}$.

answered May 14 '13 at 07:18

André Nicolas

507,029

what about $t^2$? – Ofir Attia May 14 '13 at 07:31

score 2 · Answer 2 · edited Jul 31 '17 at 01:03

2

$$\begin{align} \int^2_0 \frac{2t\, dt}{1+t} &= 2\int^2_0 \frac{1+t-1}{1+t}\,dt \\ &= 2\int^2_0 \left( \frac{1+t}{1+t}-\frac{1}{1+t}\right)\,dt \\ &= 2\int^2_0\left( 1-\frac{1}{1+t}\right)\,dt \\ &= 2[ t-\log({1+t})]_0^2 \\ &= 2[ 2-\log 3] \\ \end{align}$$

edited Jul 31 '17 at 01:03

gen-ℤ ready to perish

6,610

answered May 14 '13 at 07:53

iostream007

4,529

Integral of fractional expression $\int^3_0 \frac{dx}{1+\sqrt{x+1}}$

2 Answers2