Integrate $\int \sqrt{2x + 3} dx$

Question

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?

Answer 1

set $$t=2x+3$$ and we get $$dt=2dx$$ and you have to solve $$\int\frac{1}{2}\sqrt{t}dt$$

Answer 2

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 $u=2x+3$.

Answer 3

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.

Answer 4

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.

$\frac 13 (2x+3)^{\frac 32}+ C$

Answer 5

$$\int\sqrt{2x+3}dx=\frac{1}{2}\int(2x+3)^{\frac{1}{2}}d(2x+3)=$$ $$=\frac{1}{2}\frac{(2x+3)^{\frac{3}{2}}}{\frac{3}{2}}+C=\frac{1}{3}\sqrt{(2x+3)^3}+C.$$

Answer 6

Try

$$c(2x+3)^{3/2}.$$

The derivative is

$$c\,2\frac32(2x+1)^{1/2}$$ so that $c$ must be $\frac13$.