Evaluate $\int \frac{1}{\sqrt{x}} \, dx$ without using the power rule

Question

I am trying to evaluate $\int \frac{1}{\sqrt{x}} \, dx$ without using the power rule. I tried using integration by parts but it seems out of luck.

Multiplying both the numerator and denominator with $\sqrt{x}$ we have

$$\int \frac{1}{\sqrt{x}} \, dx = \int \frac{\sqrt{x}}{x} \, dx$$

Integrate by parts: $u = \frac{1}{x}, v' = \sqrt{x}$. We obtain the same integral as above with

$$\int \frac{\sqrt{x}}{x} \, dx = \frac{2}{3} x\sqrt{x} + \frac{2}{3} \int \frac{\sqrt{x}}{x} \, dx$$

Answer 1

for $x>0 \ $ with integration by parts $$u=\frac{1}{x} , \ \ u'=\frac{-1}{x^2} \\ v=\frac{2}{3}x^\frac{3}{2} \ \ \ \ v'=\sqrt{x}$$ $$ \begin{align} \int \frac{1}{\sqrt{x}}dx =\int \frac{\sqrt{x}}{x} dx & = \frac{2}{3}x^\frac{3}{2}.\frac{1}{x}-\int \frac{-1}{x^2}.\frac{2}{3}x^\frac{3}{2} dx +c\\ & = \frac{2}{3}\sqrt{x}+ \int \frac{2}{3} \frac{\sqrt{x}}{x} dx +c \end{align} $$ so $$ \Big(1-\frac{2}{3}\Big) \int \frac{\sqrt{x}}{x} dx = \frac{2}{3}\sqrt{x}+c $$ then $$ \int \frac{\sqrt{x}}{x} dx = 2 \sqrt{x}+c$$ with power rule $$ \begin{align} \int \frac{1}{\sqrt{x}}dx = \int x^\frac{-1}{2}dx &= \frac{1}{1-\frac{1}{2}} x^{1-\frac{1}{2}}+c \\ & =2 \sqrt{x}+c \end{align}$$

Answer 2

Evaluate $\int \frac{1}{\sqrt{x}} \; dx$ without using the power rule

Consider the $u$-substitution formula backwards by treating $u$ as the dummy variable in the integral on the right such that

$$\int \frac{1}{\sqrt{\phi(x)}} \; \phi'(x) \; dx = \int \frac{1}{\sqrt{u}} \; du$$

and if you find the antiderivative on the left with respect to $x$ then you have to undo the substitution with $x = \phi^{-1}(u)$ so that $u$ also becomes a dummy variable.

In this case, we clearly see that by choosing $\phi(u) = u^2$ then the integral on the right greatly simplifies so that we obtain

\begin{align} \int \frac{1}{\sqrt{\phi(x)}} \; \phi'(x) \; dx & = \int \frac{1}{\sqrt{x^2}}2x \; dx \\ & = 2 \int dx \\ & = 2x \Big\vert_{x = \sqrt{u}} \\ & = 2\sqrt{u} + C \end{align}

We eventually have the following result with $x$ as the dummy variable such that

$$\int \frac{1}{\sqrt{x}} \; dx = 2\sqrt{x} + C$$

This result is valid on $\mathbb{R}_{>0}$ because $\phi(u)$ is also continous and bijective on that restriction.

Note: The integral $\int dx$ is not evaluated with the power rule. If we consider $\int dx$ as the definite integral $\int^x_0 dt$ then we clearly see it corresponds to the summation of all differentials $dt$ from $0$ up to $x$, so that $\int^x_0 dt = x + C$.

Answer 3

Let $x= \sin^2 t$ so that $dx = 2\sin t \cos t \; dt.$ Then

$$\int \frac{1}{\sqrt{x}} \; dx = \int \frac{1}{\sin t} 2\sin t \cos t \; dt = 2\int \cos t \; dt$$

$$ =2\sin t + C = 2\sqrt{x} + C.$$