<s>[INST] Integrate $\int \sqrt{2x + 3} dx$ Integrate: $\int \sqrt{2x + 3} \ dx$
Doing some guesswork gives me: $\frac{1}{2}(2x + 3)^{3/2}$ to try. Differentiating this gives: $2\frac{3}{2}\frac{1}{2}(2x + 3)^{1/2}$ = $\frac{6}{4}(2x + 3)^{1/2}$
Too bad the answer should be: $\frac{1}{3}(2x + 3)^{1/2}$, so clearly I'm missing some factor of 2. Where am I going wrong?
 [/INST] Here is how you could approach this:
Your answer is within a constant factor of the real answer, so you have
$$\int \sqrt{2x+3}dx=c*(2x+3)^{\frac{3}{2}}$$
When we differentiate $c*(2x+3)^{\frac{3}{2}}$ we want to get $\sqrt{2x+3}$, so differentiate and solve for $c$:
$$\sqrt{2x+3}=\frac{d}{dx}\big(c*(2x+3)^{\frac{3}{2}}\big)$$
Use the chain rule:
$$\sqrt{2x+3}=\frac{3}{2}*2*c*\sqrt{2x+3}$$
then solve for $c$:
$$1=\frac{3}{2}*2*c$$
$$1=3c$$
$$c=\frac{1}{3}$$
That's one way to think about it.
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