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?
 A: 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.
A: set $$t=2x+3$$ and we get $$dt=2dx$$ and you have to solve $$\int\frac{1}{2}\sqrt{t}dt$$
A: You guessed the wrong constant in front, you took into account the 2 in 2x+3, but forgot to take into account the power which jumps in front.
By guesswork, you should look for an integral of the form $C (2x + 3)^{3/2}$. Differentiating you should find the constant $C$.
A non-guessing solution is to try the substitution $u=2x+3$.
A: You took a guess, and differentiated, and found out that you were off by a constant factor.  So you could stick with "guess and check" and just multiply in the factor that you are missing.
However, if you want something a little bit more systematic.
$u = 2x + 3\\
du = 2\ dx$
$\int (\sqrt {2x +3})(\frac 12) (2\ dx)\\
\int \frac 12 \sqrt {u}\ du$
Was that a factor of 2 that you missed?
Next lets rewrite it as:
$\int \frac 12 u^{\frac 12}\ du$
Integrate
$(\frac 12)(\frac 23) u^{\frac 32}+ C$
Simplify the fractions and reverse the substitution for u.
$\frac 13 (2x+3)^{\frac 32}+ C$
A: $$\int\sqrt{2x+3}dx=\frac{1}{2}\int(2x+3)^{\frac{1}{2}}d(2x+3)=$$
$$=\frac{1}{2}\frac{(2x+3)^{\frac{3}{2}}}{\frac{3}{2}}+C=\frac{1}{3}\sqrt{(2x+3)^3}+C.$$
A: Try
$$c(2x+3)^{3/2}.$$
The derivative is
$$c\,2\frac32(2x+1)^{1/2}$$ so that $c$ must be $\frac13$.
