Question

Evaluate the integral: \[ \int \sqrt{x^2 + 3x} \, dx \]

Answer 1

We are given the integral:

\[ \int \sqrt{x^2 + 3x} \, dx \]

Step 1: Complete the square

The expression inside the square root is \(x^2 + 3x\). Completing the square:

\[ x^2 + 3x = \left( x + \frac{3}{2} \right)^2 - \frac{9}{4} \]

Thus, the integral becomes:

\[ \int \sqrt{\left( x + \frac{3}{2} \right)^2 - \frac{9}{4}} \, dx \]

Step 2: Substitution

We perform the substitution \( u = x + \frac{3}{2} \), which gives \( du = dx \). The integral becomes:

\[ \int \sqrt{u^2 - \frac{9}{4}} \, du \]

Step 3: Use the standard formula for this type of integral

We use the standard integral formula for \( \int \sqrt{u^2 - a^2} \, du \), which is:

\[ \int \sqrt{u^2 - a^2} \, du = \frac{u}{2} \sqrt{u^2 - a^2} - \frac{a^2}{2} \ln \left| u + \sqrt{u^2 - a^2} \right| + C \]

Substituting \( a = \frac{3}{2} \) into this formula, we get:

\[ \frac{u}{2} \sqrt{u^2 - \frac{9}{4}} - \frac{\frac{9}{4}}{2} \ln \left| u + \sqrt{u^2 - \frac{9}{4}} \right| + C \]

Step 4: Back-substitute \( u = x + \frac{3}{2} \)

Now, substitute back \( u = x + \frac{3}{2} \) into the result:

\[ \frac{x + \frac{3}{2}}{2} \sqrt{\left( x + \frac{3}{2} \right)^2 - \frac{9}{4}} - \frac{9}{8} \ln \left| x + \frac{3}{2} + \sqrt{\left( x + \frac{3}{2} \right)^2 - \frac{9}{4}} \right| + C \]

Conclusion:

The final solution is:

\[ \int \sqrt{x^2 + 3x} \, dx = \frac{x + \frac{3}{2}}{2} \sqrt{x^2 + 3x} - \frac{9}{8} \ln \left| x + \frac{3}{2} + \sqrt{x^2 + 3x + \frac{9}{4}} \right| + C \]