To solve the integral \(\int x\sqrt{3x+1}\ dx\), we can use the method of substitution. Let's go through the steps:

### Step 1: Substitution
Let \( u = 3x + 1 \). Then, differentiate \( u \) with respect to \( x \):
\[
\frac{du}{dx} = 3 \quad \text{or} \quad dx = \frac{du}{3}
\]
Also, express \( x \) in terms of \( u \):
\[
x = \frac{u - 1}{3}
\]

### Step 2: Substitute into the integral
Substitute \( u = 3x + 1 \) and \( dx = \frac{du}{3} \) into the integral:
\[
\int x\sqrt{3x+1}\ dx = \int \left(\frac{u - 1}{3}\right)\sqrt{u}\ \frac{du}{3}
\]
Simplify the expression:
\[
\int \frac{(u - 1)\sqrt{u}}{9}\ du = \frac{1}{9} \int (u - 1)\sqrt{u}\ du
\]

### Step 3: Expand the integrand
Expand the integrand:
\[
\frac{1}{9} \int (u^{3/2} - u^{1/2})\ du
\]

### Step 4: Integrate each term
Now, integrate each term separately:
\[
\frac{1}{9} \left[\int u^{3/2}\ du - \int u^{1/2}\ du\right]
\]

For the first integral:
\[
\int u^{3/2}\ du = \frac{u^{5/2}}{\frac{5}{2}} = \frac{2}{5} u^{5/2}
\]

For the second integral:
\[
\int u^{1/2}\ du = \frac{u^{3/2}}{\frac{3}{2}} = \frac{2}{3} u^{3/2}
\]

### Step 5: Combine the results
Combine the results and factor out constants:
\[
\frac{1}{9} \left[\frac{2}{5} u^{5/2} - \frac{2}{3} u^{3/2}\right]
= \frac{2}{45} u^{5/2} - \frac{2}{27} u^{3/2}
\]

### Step 6: Substitute back \( u = 3x + 1 \)
Finally, substitute back \( u = 3x + 1 \):
\[
\frac{2}{45} (3x + 1)^{5/2} - \frac{2}{27} (3x + 1)^{3/2} + C
\]
where \( C \) is the constant of integration.

Thus, the solution to the integral is:
\[
\boxed{\int x\sqrt{3x+1}\ dx = \frac{2}{45} (3x + 1)^{5/2} - \frac{2}{27} (3x + 1)^{3/2} + C}
\]

Would you like more details or have any questions?

Here are some related problems you can try:
1. \(\int x^2\sqrt{3x+1}\ dx\)
2. \(\int x\sqrt{2x+5}\ dx\)
3. \(\int x\sqrt{4x+3}\ dx\)
4. \(\int \sqrt{5x+2}\ dx\)
5. \(\int x^3\sqrt{3x+1}\ dx\)

**Tip:** For integrals involving products of polynomials and square roots, substitution is often a powerful technique.

Learn how to solve the integral of x times square root of 3x plus 1 using substitution method with a detailed step-by-step solution. This problem covers key concepts in calculus, including integration by substitution and solving integrals involving square roots. Suitable for advanced high school students and above.

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