# Integrate $\frac{\mathrm d x}{\sqrt{2x+1}}$ from 0 to 4

$$\int_0^4 \frac{\mathrm d x}{\sqrt{2x+1}}$$

Then:

• $t=2x+1$
• $\mathrm dt=2 \mathrm dx$
• $\mathrm dx=\mathrm dt /2$

$$\int_0^4 \frac{\mathrm d t}{2\sqrt t}$$

And what next?

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Write ${1\over 2\sqrt t}={1\over 2} t^{-1/2}$. You also need to change your limits of integration... – David Mitra Jan 20 '13 at 17:37

You have done the "hard" part which is seeing the substitution, now everything you have to do is solve that integral which is really easy if you write it this way: $$1/2\int_a^b t^{-1/2}dt$$ Be carefull with the limits: $$x=0\Rightarrow t=2x+1=1$$ $$x=4\Rightarrow t=2\cdot 4+1=9$$ The integral will be: $$\frac{1}{2}\int_1^9t^{-\frac{1}{2}}dt=\frac{1}{2}\left.\frac{t^\frac{1}{2}}{\frac{1}{2}}\right|_{1}^{9}=9^{1/2}-1^{1/2}=3-1=2$$

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The integral is straightforward $$\int_0^4 \frac{\mathrm d x}{\sqrt{2x+1}}=\int_0^4 \left(\sqrt{2x+1}\,\right)^\prime \,\mathrm d x= \left(\sqrt{2x+1}\,\right)\bigg|_0^4=3-1=2.$$

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$$\int_0^4 \frac{dx}{\sqrt{2x+1}} = \frac{1}{2}\int_{1}^{9} \frac{du}{\sqrt{u}} = \frac{1}{2}\int_{1}^{9} u^{-1/2}\, du$$

substituting $u=2x+1$, as you have found (however, note the new limits are $2\cdot 0 +1 = 1$ and $2\cdot 4 +1=9$). Now use

$$\int x^\alpha\, dx = \frac{x^{\alpha+1}}{\alpha+1}+C,\forall \alpha\neq -1$$

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