If $$ y=x \sqrt{2 x+1} $$, show that $$ \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{3 x+1}{\sqrt{2 x+1}} $$ Hence show that $$ \int_{0}^{4} \frac{3 x+1}{2 \sqrt{2 x+1}} \mathrm{~d} x=6 $$ and $$ \int_{0}^{4} \frac{6 x}{\sqrt{2 x+1}} \mathrm{~d} \boldsymbol{x}=20 $$

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We start with
$$
y = x \sqrt{2x+1}.
$$

1. **Differentiate $$ y $$ with respect to $$ x $$:**

Write $$ y $$ as a product of two functions:
   $$
   u(x)=x \quad \text{and} \quad v(x)=\sqrt{2x+1}.
   $$
   
   The derivative $$ u'(x)=1 $$. For $$ v(x) $$, we use the chain rule:
   $$
   v'(x) = \frac{d}{dx}\sqrt{2x+1} = \frac{1}{2\sqrt{2x+1}} \cdot \frac{d}{dx}(2x+1) = \frac{1}{\sqrt{2x+1}}.
   $$
   
   Now, applying the product rule:
   $$
   \frac{dy}{dx} = u'(x)v(x) + u(x)v'(x) = \sqrt{2x+1} + x\left(\frac{1}{\sqrt{2x+1}}\right).
   $$
   
   Combine the terms using a common denominator:
   $$
   \frac{dy}{dx} = \frac{(2x+1)+x}{\sqrt{2x+1}} = \frac{3x+1}{\sqrt{2x+1}}.
   $$

2. **First Integral: Show that**
   $$
   \int_{0}^{4} \frac{3x+1}{2\sqrt{2x+1}}\,dx = 6.
   $$

Rewrite the integral by factoring out the constant:
   $$
   I_1 = \frac{1}{2}\int_{0}^{4} \frac{3x+1}{\sqrt{2x+1}}\,dx.
   $$

Use the substitution:
   $$
   u = 2x+1 \quad \Rightarrow \quad du = 2\,dx \quad \Rightarrow \quad dx = \frac{du}{2}.
   $$
   When $$ x=0 $$, $$ u=1 $$; and when $$ x=4 $$, $$ u=9 $$.

Also, express $$ x $$ in terms of $$ u $$:
   $$
   x = \frac{u-1}{2}.
   $$

Substitute into the integrand. Notice:
   $$
   3x+1 = 3\left(\frac{u-1}{2}\right)+1 = \frac{3u-3+2}{2} = \frac{3u-1}{2}.
   $$
   
   The integral becomes:
   $$
   I_1 = \frac{1}{2} \int_{u=1}^{9} \frac{\frac{3u-1}{2}}{\sqrt{u}} \cdot \frac{du}{2} 
   = \frac{1}{8} \int_{1}^{9} \frac{3u-1}{\sqrt{u}}\,du.
   $$
   
   Simplify the integrand:
   $$
   \frac{3u-1}{\sqrt{u}} = 3\sqrt{u} - \frac{1}{\sqrt{u}}.
   $$
   
   Now, the integral is:
   $$
   I_1 = \frac{1}{8} \left[\int_{1}^{9} 3\sqrt{u}\,du - \int_{1}^{9} \frac{1}{\sqrt{u}}\,du \right].
   $$
   
   Evaluate each term:
   
   - For $$\int 3\sqrt{u}\,du$$:
     $$
     \int 3\sqrt{u}\,du = 3\int u^{\frac{1}{2}}\,du = 3\cdot \frac{2}{3}u^{\frac{3}{2}} = 2u^{\frac{3}{2}}.
     $$
   
   - For $$\int \frac{1}{\sqrt{u}}\,du$$:
     $$
     \int \frac{1}{\sqrt{u}}\,du = \int u^{-\frac{1}{2}}\,du = 2u^{\frac{1}{2}}.
     $$
   
   Combine the results:
   $$
   I_1 = \frac{1}{8} \left[2u^{\frac{3}{2}} - 2u^{\frac{1}{2}}\right]_{1}^{9} 
   = \frac{1}{4} \left[u^{\frac{3}{2}} - u^{\frac{1}{2}}\right]_{1}^{9}.
   $$
   
   Evaluate at the limits:
   $$
   \left[u^{\frac{3}{2}} - u^{\frac{1}{2}}\right]_{1}^{9} 
   = \left(9^{\frac{3}{2}} - 9^{\frac{1}{2}}\right) - \left(1^{\frac{3}{2}} - 1^{\frac{1}{2}}\right).
   $$
   
   Recall:
   $$
   9^{\frac{1}{2}}=3, \quad 9^{\frac{3}{2}}=27,\quad 1^{\frac{1}{2}}=1,\quad 1^{\frac{3}{2}}=1.
   $$
   
   So,
   $$
   \left(27-3\right) - \left(1-1\right)=24.
   $$
   
   Hence,
   $$
   I_1 = \frac{1}{4}(24)=6.
   $$

3. **Second Integral: Show that**
   $$
   \int_{0}^{4} \frac{6x}{\sqrt{2x+1}}\,dx = 20.
   $$

Use the substitution:
   $$
   u = 2x+1 \quad \Rightarrow \quad du = 2\,dx \quad \Rightarrow \quad dx = \frac{du}{2}.
   $$
   When $$ x=0 $$, $$ u=1 $$; and when $$ x=4 $$, $$ u=9 $$.

Express $$ x $$ in terms of $$ u $$:
   $$
   x = \frac{u-1}{2}.
   $$

Substitute into the integrand:
   $$
   6x = 6\left(\frac{u-1}{2}\right)=3(u-1).
   $$
   
   The integrand becomes:
   $$
   \frac{6x}{\sqrt{2x+1}}\,dx = \frac{3(u-1)}{\sqrt{u}} \cdot \frac{du}{2} = \frac{3}{2}\cdot \frac{u-1}{\sqrt{u}}\,du.
   $$
   
   Rewrite the fraction:
   $$
   \frac{u-1}{\sqrt{u}} = \sqrt{u} - \frac{1}{\sqrt{u}}.
   $$
   
   Therefore, the integral is:
   $$
   I_2 = \frac{3}{2} \int_{1}^{9} \left(\sqrt{u} - \frac{1}{\sqrt{u}}\right)\,du.
   $$
   
   Now, integrate term by term:
   
   - For $$\int \sqrt{u}\,du$$:
     $$
     \int \sqrt{u}\,du = \int u^{\frac{1}{2}}\,du = \frac{2}{3}u^{\frac{3}{2}}.
     $$
   
   - For $$\int \frac{1}{\sqrt{u}}\,du$$:
     $$
     \int \frac{1}{\sqrt{u}}\,du = \int u^{-\frac{1}{2}}\,du = 2u^{\frac{1}{2}}.
     $$
   
   Hence,
   $$
   I_2 = \frac{3}{2} \left[\frac{2}{3}u^{\frac{3}{2}} - 2u^{\frac{1}{2}}\right]_{1}^{9}.
   $$
   
   Simplify the constant factors:
   $$
   \frac{3}{2} \cdot \frac{2}{3} =1 \quad \text{and} \quad \frac{3}{2}\cdot 2 =3.
   $$
   
   This gives:
   $$
   I_2 = \left[u^{\frac{3}{2}} - 3u^{\frac{1}{2}}\right]_{1}^{9}.
   $$
   
   Evaluate at the limits:
   $$
   \left(9^{\frac{3}{2}} - 3\cdot 9^{\frac{1}{2}}\right) - \left(1^{\frac{3}{2}} - 3\cdot 1^{\frac{1}{2}}\right)
   = \left(27-3\cdot 3\right) - \left(1-3\right)
   = \left(27-9\right) - \left(-2\right)
   = 18 + 2 = 20.
   $$

Thus, the final results are:
$$
\frac{dy}{dx}=\frac{3x+1}{\sqrt{2x+1}},\quad \int_{0}^{4} \frac{3x+1}{2\sqrt{2x+1}}\,dx = 6,\quad \text{and}\quad \int_{0}^{4} \frac{6x}{\sqrt{2x+1}}\,dx = 20.
$$
$$ \frac{dy}{dx} = \frac{3x + 1}{\sqrt{2x + 1}}, \quad \int_{0}^{4} \frac{3x + 1}{2\sqrt{2x + 1}}\,dx = 6, \quad \int_{0}^{4} \frac{6x}{\sqrt{2x + 1}}\,dx = 20 $$

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