The integral of the root x, can be found using power rule. The integral of $\sqrt x $ with respect to (x) is:

\[

\int \sqrt x \, dx =\frac{2}{3} x^{3/2} + C

\]

Here, (C) represents the constant of integration, which is added because the process of integration determines the antiderivative up to an arbitrary constant.

## Proof of integration of root x

To integrate the square root of $x $ (i.e., $\sqrt{x}$), we express it as $x^{1/2} $. The integral is then:

$$ \int \sqrt{x} \, dx = \int x^{1/2} \, dx $$

To integrate this, we use the power rule for integration:

$$ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C $$

where $C $ is the constant of integration. Applying this rule:

$$ \int x^{1/2} \, dx = \frac{x^{(1/2) + 1}}{(1/2) + 1} + C = \frac{x^{3/2}}{3/2} + C $$

Simplifying the fraction:

$$ = \frac{2}{3} x^{3/2} + C $$

So, the integral of (\sqrt{x}$ is:

$$ \frac{2}{3} x^{3/2} + C $$

## Integration based on root x

$$ \int \sqrt{ax + b} \, dx = \frac{2}{3a} (ax + b)^{3/2} + C $$

**Proof**

Choose a substitution that simplifies the integrand. Let ( $u = ax + b$ ). Then, ( $du = a \, dx$ ) or ( $dx = \frac{du}{a}$ ).

Substitute ( u ) and ( dx ) into the integral:

$$= \int \sqrt{u} \cdot \frac{du}{a} $$

$$= \frac{1}{a} \int u^{1/2} \, du $$

$$= \frac{1}{a} \cdot \frac{2}{3} u^{3/2} + C $$

Here, ( C ) is the constant of integration.

Substitute back for ( u ) to get the final answer:

$$\int \sqrt{ax + b} \, dx= \frac{2}{3a} (ax + b)^{3/2} + C $$

## Definite Integral of root x

To evaluate a definite integral of the root x, $\int_a^b \sqrt x \, dx$, where $ a $ and $ b $ are the limits of integration, we follow a similar process as with the indefinite integral and the we’ll apply the limits at the end.

$$

\int_a^b \sqrt(x) \, dx = [ \frac{2}{3} x^{3/2}]_a^b

$$

## Solved Examples

**Question 1**

$$ \int \sqrt{3x – 1} \, dx $$

**Solution**

Choose a substitution that simplifies the integrand. Let $ u = 3x – 1 $. Then, $ du = 3 \, dx $ or $ dx = \frac{du}{3} $.

Substitute $ u $ and $ dx $ into the integral:

$$ =\int \sqrt{u} \cdot \frac{du}{3} $$

You can simplify this to:

$$ =\frac{1}{3} \int u^{1/2} \, du $$

Now integrate with respect to $ u $:

$$ =\frac{1}{3} \cdot \frac{2}{3} u^{3/2} + C $$

Here, $ C $ is the constant of integration.

Substitute back for $ u $ to get the final answer:

$$= \frac{2}{9} (3x – 1)^{3/2} + C $$

**Question 2**

$$ \int \sqrt{x} \cdot \ln(x) \, dx $$

**Solution**

Integrating $ \sqrt{x} \cdot \ln(x) $ requires the use of integration by parts.

$$ \int u \, dv = uv – \int v \, du $$

- Let $ u = \ln(x) $ (since the derivative of $ \ln(x) $ is simpler than its integral).
- Then $ dv = \sqrt{x} \, dx $ (since we know how to integrate $ \sqrt{x} $).

Then

- $ du = \frac{1}{x} \, dx $.
- $ v = \int \sqrt{x} \, dx = \int x^{1/2} \, dx = \frac{2}{3} x^{3/2} $ (using the power rule).

Therefore

$$ \int \sqrt{x} \cdot \ln(x) \, dx = uv – \int v \, du $$

$$ = \ln(x) \cdot \frac{2}{3} x^{3/2} – \int \frac{2}{3} x^{3/2} \cdot \frac{1}{x} \, dx $$

$$ = \frac{2}{3} x^{3/2} \ln(x) – \frac{2}{3} \int x^{1/2} \, dx $$

$$ = \frac{2}{3} x^{3/2} \ln(x) – \frac{2}{3} \cdot \frac{2}{3} x^{3/2} + C $$

$$ = \frac{2}{3} x^{3/2} \ln(x) – \frac{4}{9} x^{3/2} + C $$

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